Chapter 23: Circle and System of Circles - PDF

60 80 Circle and System of Circles 4.1 Definition. E3 A circle is defined as the locus of a point which moves in a plane such that its distance from a fixed point in that plane always remains the same i.e. (Moving constant. point) P The fixed point is called the centre of the circle and the fixed di...

60 80 Circle and System of Circles 4.1 Definition. E3 A circle is defined as the locus of a point which moves in a plane such that its distance from a fixed point in that plane always remains the same i.e. (Moving constant. point) P The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle. Note :  O Q If r(r  0) is the radius of a circle, the diameter R Fixed point Plane ID d  2r is the maximum distance between any two points on the given circle.  The length of the curve or perimeter (also called circumference) of circle  2r or d. d 2. 4 U  The area of circle  r 2 or  Line joining any two points of a circle is called chord of circle. D YG  Curved section between any two points of a circle is called arc of circle.  Angle subtended at the centre of a circle by any arc = arc/radius.  Angle subtended at the centre of a circle by an arc is double of angle subtended at the circumference of a circle. 4.2 Standard forms of Equation of a Circle. (1) General equation of a circle : The general equation of a circle is x 2  y 2  2 gx  2 fy  c  0, U where g, f, c are constant. (i) Centre of the circle is (–g, –f). i.e., (  ST (ii) Radius of the circle is Note 1 1 coefficient of x,  coefficient of y) 2 2 g2  f 2  c. : The general equation of second degree ax 2  by 2  2hxy  2 gx  2 fy  c  0 represents a circle if a  b  0 and h  0.  Locus of a point P represent a circle if its distance from two points A and B is not equal i.e. PA  kPB represent a circle if k  1.  Discussion on nature of the circle :  If g 2  f 2  c  0 , then the radius of the circle will be real. Hence, in this case, it is possible to draw a circle on a plane.  If g 2  f 2  c  0 , then the radius of the circle will be zero. Such a circle is known as point circle. Circle and System of Circles 81 g 2  f 2  c of the circle will be an imaginary If g 2  f 2  c  0 , then the radius  number. Hence, in this case, it is not possible to draw a circle.  Special features of the general equation x 2  y 2  2 gx  2 fy  c  0 of the circle : 60 This equation has the following peculiarities :  It is a quadratic equation in x and y. Here the co-efficient of x 2  the co-efficient of y 2  In working out problems it is advisable to keep the co-efficient of x 2 and y 2 as unity. There is no term containing xy, i.e. the co-efficient of the term xy is zero.  This equation contains three arbitrary constants. If we want to find the equation of a circle of which neither the centre nor the radius is known, we take the equation in the above form and determine the values of the constants g, f, c for the circle in question from the given geometrical conditions.  Keeping in mind the above special features, we can say that the equation ID E3  ax 2  ay 2  2 gx  2 fy  c  0 …..(i) also represents a circle. U This equation can also be written as x 2  y 2  2 g f c x  2 y   0, dividing by a  0. a a a D YG g2 f 2 c  g  f  Hence, the centre     ,  and radius  a2 a2 a  a a  (2) Central form of equation of a circle : The equation of a circle having centre (h, k) and radius r is (x  h)2  (y  k )2  r 2 Note :  If the centre is origin, then the equation of the circle is x y r 2 P(x,y) (h,k C ) 2 U 2 r If r = 0, then circle is called point circle and its equation is (x  h)2  (y  k )2  0  ST (3) Concentric circle : Two circles having the same centre C (h, k) but different radii r1 and r2 respectively are (x  h)  (y  k )  2 2 called r22 , r1 concentric circles. Thus the circles (x  h)2  (y  k )2  r12 and  r2 are concentric circles. Therefore, the equations of concentric circles differ only in constant terms. (4) Circle on a given diameter : The equation of the circle drawn on the straight line joining ( x 1 , y 1 ) and (x 2 , y 2 ) two given points as diameter is P(x,y) (x  x 1 )(x  x 2 )  (y  y 1 )(y  y 2 )  0. Note : 90° r If the coordinates of the end points of a diameter of a circle are given, we can also find the equation of the circle by finding the coordinates of (x1,y1) A C (h,k ) B(x2,y2 ) 82 Circle and System of Circles the centre and radius. The centre is the mid-point of the diameter and radius is half of the length of the diameter. (5) Parametric coordinates (i) The parametric coordinates of any point on the circle (x  h)2  (y  k )2  r 2 are given by (0    2 ) In particular, co-ordinates of any point on the circle x 2  y 2  r 2 are (r cos  , r sin  ) , (0    2 ) (ii) The parametric co-ordinates x  y  2 gx  2 fy  c  0 are of any 2 point on the circle E3 2 60 (h  r cos  , k  r sin  ) , x   g  (g 2  f 2  c ) cos  and y   f  (g 2  f 2  c) sin  , (0    2 ) (6) Equation of a circle under given conditions: The general equation of circle, i.e., x  y 2  2 gx  2 fy  c  0 contains three independent constants g, f and c. Hence for determining 2 ID the equation of a circle, three conditions are required. (i) The equation of the circle through three non-collinear points A (x 1 , y 1 ), B (x 2 , y 2 ), C (x 3 , y 3 ): Let the equation of circle be x 2  y 2  2 gx  2 fy  c  0 …..(i) equation. Hence solving equations U If three points (x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ) lie on the circle (i), their co-ordinates must satisfy its x 12  y 12  2 gx1  2 fy1  c  0 …..(ii) …..(iii) x 32  y 32  2 gx 3  2 fy3  c  0 …..(iv) D YG x 22  y 22  2 gx 2  2 fy2  c  0 g, f, c are obtained from (ii), (iii) and (iv). Then to find the circle (i). Alternative method (1) The equation of the circle through three non-collinear A (x 1 , y 1 ), B (x 2 , y 2 ), C (x 3 , y 3 ) is ST U x2 x 12 x 22 x 32  y2  y 12  y 22  y 32 x x1 x2 x3 y y1 y2 y3 points 1 1 0 1 1 (2) From given three points taking any two as extremities of diameter of a circle S = 0 and equation of straight line passing through these two points is L = 0. Then required equation of circle is S  L  0 , where  is a parameter which can be found out by putting third point in the equation. Note :  Cyclic quadrilateral : If all the four vertices of a quadrilateral lie on a circle, then the quadrilateral is called a cyclic quadrilateral. The four vertices are said to be concylic. 4.3 Equation of a Circle in Some special cases. (1) If centre of the circle is (h, k ) and it passes through origin then its equation is (x  h)2  (y  k )2  h 2  k 2  x 2  y 2  2hx  2ky  0 Circle and System of Circles 83 (2) If the circle touches x axis then its equation is (Four cases) (x  h)2  (y  k )2  k 2 Y (h,k) k k (–h,– k) k (h,–k) X 60 X (– h,k)k Y If the circle touches y axis then its equation is (Four cases) (x  h)2  (y  k )2  h 2 (3) X h h (h,k) (– h,k) X ID h h (–h,– (h,–k) k) Y E3 Y Y (4) If the circle touches both the axes then its equation is (Four cases) D YG U (x  r)2  (y  r)2  r 2 X X (–r,– (r,–r) r) Y (5) If the circle touches x- axis at origin (Two cases) Y 2 U  x 2  y 2  2ky  0 (0,k ) 2 X X (0,– k) x  (y  k )  k 2 (r,r) (– r,r) Y Y (6) If the circle touches y-axis at origin (Two cases) ST (x  h)2  y 2  h 2  x  y  2 xh  0 2 2 X (– h,0) (h,0 ) X Y (7) If the circle passes through origin and cut intercepts of a and b on axes, the equation of circle is (Four cases) x 2  y 2  ax  by  0 and centre is (a / 2, b / 2) Y b a X 84 Circle and System of Circles Note :  Circumcircle of a triangle : If we are given sides of a triangle, then first we should find vertices then we can find the equation of the circle using general form. Alternate : If equation of the sides are L1  0, L 2  0 and L 3  0 , then equation of circle is (L1. L 2 )   (L 2. L 3 )  (L 3.L1 )  0 , where  and  are the constant which can If the circle x 2  y 2  2 gx  2 fy  c  0 touches X-axis, then  f  g 2  f 2  c or U  ID E3 60 be found out by the conditions, coefficient of x 2  coefficient of y 2 and coefficient of xy = 0  If the triangle is right angled then its hypotenuse is the diameter of the circle. So using diameter form we can find the equation.  Circumcircle of a square or a rectangle : Diagonals of the square and rectangle will be diameters of the circumcircle. Hence finding the vertices of a diagonal, we can easily determine the required equation. Alternate : If sides of a quadrilateral are L1  0, L 2  0, L 3  0 and L 4  0. Then equation of circle is L1 L 3  L 2 L 4  0, where  is a constant which can be obtained by the condition of circle.  If a circle is passing through origin then constant term is absent i.e. x 2  y 2  2 gx  2 fy  0 g2  c  If the circle x 2  y 2  2 gx  2 fy  c  0 touches Y-axis, then  g  g 2  f 2  c or  D YG f2  c If the circle x 2  y 2  2 gx  2 fy  c  0 touches both axes, then  g   f  g 2  f 2  c or g2 = f 2 = c Example: 1 A point P moves in such a way that the ratio of its distances from two coplanar points is always fixed number ( 1). Then its locus is [IIT 1970] (a) Straight line Solution: (b) (b) Circle U Under given conditions, we get   1) (c) Parabola (d) A pair of straight lines Let two coplanar points are (0, 0) and (a, 0) and coordinates of point P is (x, y). x 2  y2 (x  a)2  y 2  (where  is any number and ST  2   (a 2  2 ax )  0 , which is equation of a circle.  x 2  y 2  2 [(x  a)2  y 2 ]  x 2  y 2   2    1  Example : 2 The lines 2 x  3 y  5 and 3 x  4 y  7 are the diameters of a circle of area 154 square units. The equation of the circle is [IIT 1989; AIEEE 2003; DCE 2001] (a) x  y  2 x  2y  62 (b) x  y  2 x  2y  47 (c) x  y  2 x  2y  47 (d) x 2  y 2  2 x  2y  62 2 2 2 2 2 2 Solution : (b) Centre of circle = Point of intersection of diameters, On solving equations, 2 x  3 y  5 and 3 x  4 y  7 , we get, (x , y )  (1,1)  Centre of circle = (1,  1). Now area of circle = 154  r 2  154  r = 7 Hence, the equation of required circle is (x  1)2  (y  1)2  (7)2  x 2  y 2  2 x  2 y  47. Circle and System of Circles 85 Example : 3 The equation of a circle with origin as centre passing through the vertices of an equilateral triangle whose median is of length 3a is [BIT Ranchi 1992; AIEEE 2002] (a) x 2  y 2  9a 2 (b) x 2  y 2  16a 2 (c) x 2  y 2  a2 (d) None of these 60 Solution : (d) Since the triangle is equilateral, therefore centroid of the triangle is the same as the circumcentre 2 2 and radius of the circum-circle  (median)  (3 a)  2 a divides [ Centroid 3 3 median in ratio of 2 : 1] Hence, the equation of the circum-circle whose centre is (0, 0) and radius 2a is x 2  y 2  (2a) 2  x 2  y 2  4a 2 A circle of radius 5 units touches both the axes and lies in first quadrant. If the circle makes one complete roll on x-axis along the positive direction of x-axis, then its equation in the new position is E3 Example : 4 (a) x 2  y 2  20x  10 y  100  2  0 (b) x 2  y 2  20x  10 y  100  2  0 (c) x 2  y 2  20x  10 y  100  2  0 (d) None of these Solution : (d) The x-coordinate of the new position of the circle is 5 + circumferrence of the first circle  5  10  ID The y-coordinate is 5 and the radius is also 5. Hence, the equation of the circle in the new position is (x  5  10 ) 2  (y  5) 2  (5) 2  x 2  25  100  2  10 x  100   20 x  y 2  25  10 y  25 Example : 5 U  x 2  y 2  20 x  10 x  10 y  100  2  100   25  0 The abscissae of A and B are the roots of the equation x 2  2ax  b 2  0 and their ordinates are the roots of the equation y 2  2by  q 2  0. The equation of the circle with AB as diameter is D YG [IIT 1984] (a) x  y  2ax  2by  b  q  0 (b) x 2  y 2  2ax  by  b 2  q 2  0 (c) x 2  y 2  2ax  2by  b 2  q 2  0 (d) None of these 2 2 2 2 Solution : (a) Let x 1 , x 2 and y 1 , y 2 be roots of x 2  2ax  b 2  0 and y 2  2by  q 2  0 respectively. Then, x 1  x 2  2a, x 1 x 2  b 2 and y 1  y 2  2b, y 1 y 2  q 2 The equation of the circle with A (x 1 , y 1 ) and B (x 2 , y 2 ) as the end points of diameter is (x  x 1 ) (x  x 2 )  (y  y1 ) (y  y 2 )  0 U x 2  y 2  x (x 1  x 2 )  y(y 1  y 2 )  x 1 x 2  y 1 y 2  0 ; The equation of a circle of radius 1 touching the circles x 2  y 2  2| x |  0 is (a) x 2  y 2  2 3 x  2  0 (b) x 2  y 2  2 3 y  2  0 (c) x 2  y 2  2 3 y  2  0 (d) x 2  y 2  2 3 x  2  0 ST Example : 6 x 2  y 2  2ax  2by  b 2  q 2  0 Solution : (b,c) The given circles are x 2  y 2  2 x  0, x  0, and x 2  y 2  2 x  0, x  0. From the figure, the centres of the required circles will be (0, 3 ) and (0,  3 ). 1 1 (– 1,0)  The equations of the circles are (x  0) 2  (y  3 ) 2  1 2.  x 2  y2  3  2 3y  1  x 2  y2  2 3y  2  0 Example : 7 If the line x  2by  7  0 is a diameter of the circle x 2  y 2  6 x  2y  0, then b = [MP PET 1991] (a) 3 (b) – 5 (c) – 1 1 (d) 5 1 1 1 (1,0 ) 86 Circle and System of Circles Solution : (d) Here the centre of circle (3, –1) must lie on the line x  2by  7  0. Therefore, 3  2b  7  0  b  5 Example : 8 The centre of the circle r 2  2  4 r cos   6 r sin  is (a) (2, 3) (b) (– 2, 3) (c) (– 2, – 3) (d) (2, – 3) Solution : (b) Let r cos  = x and r sin  = y Squaring and adding, we get r 2  x 2  y 2.  x 2  y 2  4 x  6y  2  0 60 Putting these values in given equation, x 2  y 2  2  4 x  6 y Hence, centre of the circle = (–2, 3) Example : 9 The number of integral values of  for which x 2  y 2  x  (1  )y  5  0 is the equation of a circle (a) 14 E3 whose radius cannot exceed 5, is (b) 18 (c) 16 (d) None of these 2 2 (1   )     1    Solution : (c) Centre of circle    ,   5 5  ; Radius of circle      2 2   2  2  2 2  2  119  0 , 1  239 1  239   2 2    7.2    8.2 (Nearly). ID      7,  6,................,7, 8. Hence number of integral values of  is 16. Let f (x, y)  0 be the equation of a circle. If f (0, )  0 has equal roots   2, 2 and f (, 0)  0 has roots  U Example : 10 4 ,5, then the centre of the circle is 5  29  (a)  2,   10  D YG  29  , 2 (b)   10  29   (c)   2,  10   (d) None of these Solution : (b)  f (x , y )  x 2  y 2  2 gx  2 fy  c  0 Now, f (0,  )   2  2 f  c  0 and its roots are 2, 2.  2  2  2 f , 2  2  c, f (, 0)   2  2 g  c  0, and its roots are  i.e., g  4 , 5. 5 29  29  , 2. , c  4. Hence, centre of the circle  ( g,  f )   10  10  If the lines 3 x  4 y  4  0 and 6 x  8 y  7  0 are tangents to a circle, then the radius of the circle is U Example : 11 4 4  5  2 g,  5  c, 5 5 i.e. f  2, c  4 (b) 3/4 ST (a) 3/2 [IIT 1984; MP PET 1994, 2002; Rajasthan PET 1995, 97; Kurukshetra CEE 1998] (c) 1/10 (d) 1/20 Solution : (b) Since both tangents are parallel to each other. The diameter of the circle is perpendicular distance 7 between the parallel lines (tangents) 3 x  4 y  4  0 and 3 x  4 y   0 and so it is equal to 2 4 7/2 3  . 2 9  16 9  16 Hence radius of circle is 3. 4 (0, 1) Alternative method : Perpendicular distance = i.e., 3 Diameter = 2 3(0 )  4 (1)  7 / 2 3  , 5 2 3x – 4y+4=0 3x – 4y – 7/2=0 Circle and System of Circles 87 Hence radius of circle is 3. 4 4.4 Intercepts on the Axes. The lengths of intercepts made by the circle x 2  y 2  2 gx  2 fy  c  0 with X and Y axes are Let the equation of circle be x 2  y 2  2 gx  2 fy  c  0 60 2 g 2  c and 2 f 2  c respectively.......(i) Length of intercepts on x-axis and y-axis are | AB| | x 2  x 1 | and | CD| | y 2  y 1 | respectively. The circle intersects the x-axis, when y = 0, then x 2  2 gx  c  0. E3 Y Since the circle intersects the x-axis at A (x 1 , 0 ) and B (x 2 , 0). (0,y2) D Then x 1  x 2  2 g, x 1 x 2  c ID  | AB| | x 2  x 1 |  (x 2  x 1 )2  4 x 1 x 2  2 (g 2  c) As the circle intersects the y-axis, when x = 0, then y 2  2 fy  c  0 (0,y1) X C OA B (x1,0 (x2,0 ) ) Since the circle intersects the y-axis at C (0, y1) and D (0, y2), then y 1  y 2  2 f , y 1 y 2  c :   2 ( f 2  c). If g 2  c, then the roots of the equation x 2  2 gx  c  0 are real and distinct, D YG Note y 2  y 1 2  4 y 2 y 1 U  | CD| | y 2  y 1 |  so the circle x 2  y 2  2 gx  2 fy  c  0 meets the x-axis in two real and distinct points and the length of the intercept on x-axis is 2 g 2  c.  If g 2  c, then the roots of the equation x 2  2 gx  c  0 are real and equal, so the circle touches x-axis and the intercept on x-axis is zero.  If g 2  c, then the roots of the equation x 2  2 gx  c  0 are imaginary, so the circle U x 2  y 2  2 gx  2 fy  c  0 does not meet x-axis in real points.  Similarly, the circle x 2  y 2  2 gx  2 fy  c  0 cuts the y-axis in real and distinct ST points, touches or does not meet in real points according as f 2 ,  or  c. 4.5 Position of a point with respect to a Circle. A point (x 1 , y 1 ) lies outside, on or inside a circle S  x 2  y 2  2 gx  2 fy  c  0 according as S 1  x 12  y 12  2 gx1  2 fy1  c is positive, zero or negative i.e., S 1  0  Point is outside the circle. S 1  0  Point is on the circle. S 1  0  Point is inside the circle. (1) The least and greatest distance of a point from a circle : Let S = 0 be a circle and A ( x 1 , y 1 ) be a point. If the diameter of the circle is passing through the circle at P and Q, then Q r P A C 88 Circle and System of Circles AP  AC  r  least distance ; AQ  AC  r  greatest distance where 'r' is the radius and C is the centre of the circle. The number of points with integral coordinates that are interior to the circle x 2  y 2  16 is (a) 43 (b) 49 (c) 45 (d) 51 60 Example : 12 Solution : (c) The number of points is equal to the number of integral solutions (x, y) such that x 2  y 2  16. So, x, y are integers such that 3  x  3,  3  y  3 satisfying the inequation x 2  y 2  16. The number E3 of selections of values of x is 7, namely –3, –2, –1, 0, 1, 2, 3. The same is true for y. So the number of ordered pairs (x, y) is 7×7. But (3, 3), (3, –3), (–3, 3), (–3, –3) are rejected because they do not satisfy the inequation x 2  y 2  16. So the number of points is 45. Example : 13 The range of values of a for which the point (a, 4) is outside the circles x 2  y 2  10 x  0 and x 2  y 2  12 x  20  0 is (a) (, 8)  (2, 6)  (6,  ) ID (b) (– 8, – 2) (c) (, 8) (2,  ) (d) None of these Solution : (a) For circle, x  y  10 x  0 ; 2 2 For circle,  a 2  10 a  16  0  (a  8) (a  2)  0  a  8 U a 2  (4 ) 2  10 a  0 x 2  y 2  12 x  20  0 ; a 2  (4 ) 2  12 a  20  0  a  R  {6} a  2 …..(i)  a 2  12 a  36  0 …..(ii) D YG  (a  6) 2  0 or Taking common values from (i) and (ii), a  (, 8)  (2, 6)  (6,  ). 4.6 Intersection of a Line and a Circle. Let the equation of the circle be x 2  y 2  a2 and the equation of the line be y  mx  c …..(i) …..(ii) …..(iii) (1  m 2 )x 2  2mcx  c 2  a 2  0 Case I: When points of intersection are real and distinct. In this case (iii) has two distinct roots. B 2  4 AC  0 or 4 m 2 c 2  4(1  m 2 )(c 2  a 2 )  0 or a 2  ST  U From (i) and (ii), x 2  (mx  c)2  a 2 or or a | c| (1  m 2 ) c2 1  m2  length of perpendicular from (0, 0) to y  mx  c P Q O  a > length of perpendicular from (0, 0) to y  mx  c Thus, a line intersects a given circle at two distinct points if radius of circle is greater than the length of perpendicular from centre of the circle to the line. Case II: When the points of intersection are coincident in this case (iii) has two equal roots.  B 2  4 AC  0  4 m 2 c 2  4(1  m 2 )(c 2  a 2 )  0  c2 a  1  m2 2 R or a  | c| (1  m 2 ) O Circle and System of Circles 89 a = length of perpendicular from the point (0, 0) to y  mx  c.  B 2  4 AC  0  4 m 2 c 2  4(1  m 2 )(c 2  a 2 )  0 ,  a 2  or a | c|  length of perpendicular from (0, 0) to y  mx  c O E3 (1  m 2 ) c2 1m2 60 Thus, a line touches the circle if radius of circle is equal to the length of perpendicular from centre of the circle to the line. Case III: When the points of intersection are imaginary. In this case (iii) has imaginary roots.  a < length of perpendicular from (0, 0) to y  mx  c ID Thus, a line does not intersect a circle if the radius of circle is less than the length of perpendicular from centre of the circle to the line. (1) The length of the intercept cut off from a line by a circle : The length of the intercept cut off a 2 (1  m 2 )  c 2 1m2 (2) Condition of tangency : A line L = 0 touches the circle S = 0, if length of perpendicular drawn from the centre of the circle to the line is equal to radius of the circle i.e. p = r. This is the condition of tangency for the line L = 0. U from the line y  mx  c by the circle x 2  y 2  a 2 is 2 D YG Circle x 2  y 2  a 2 will touch the line y  mx  c if c  a 1  m 2 Again, (i) If a 2 (1  m 2 )  c 2  0 line will meet the circle at real and different points. (ii) If c 2  a 2 (1  m 2 ) line will touch the circle. (iii) If a 2 (1  m 2 )  c 2  0 line will meet circle at two imaginary points. Example : 14 If the straight line y  mx is outside the circle x 2  y 2  20 y  90  0, then [Roorkee 1999] U (a) m > 3 (b) m < 3 (c) | m | > 3 (d) | m | < 3 Solution : (d) If the straight line y  mx is outside the given circle then perpendicular distance of line from centre of circle > radius of circle 10  (1  m 2 )  10  m 2  9  | m|  3  10 1m2 If the chord y  mx  1 of the circle x 2  y 2  1 subtends an angle of measure 45o at the major segment of the circle then value of m is ST Example : 15 [AIEEE 2002] (a) 2 (b) – 2 (c)  1 (d) None of these Solution : (c) Given circle is x 2  y 2  1, C (0,0) and radius = 1 and chord is y  mx  1 CP ; CP = Perpendicular distance from (0, 0) to chord y  mx  1 CR 1 (CR = radius =1) 2 m 1 cos 45 o  CP  1 cos 45 o  m2  1  1  1 2 1 m2  1 4.7 Tangent to a Circle at a given Point.  m 2  1  2  m  1. 45 ° C(0,0 ) 1 45 45 ° ° P y=mx+1 R 90 Circle and System of Circles The limiting position of the line PQ, when Q moves towards P and ultimately coincides with P, is called the tangent to the circle at the point P. The point P is called the point of contact. (1) Point form (i) The equation of tangent at (x1, y1) to circle x 2  y 2  a 2 is xx 1  yy 1  a 2 (ii) The equation of tangent at (x 1 , y 1 ) to circle x 2  y 2  2 gx  2 fy  c  0 is  For equation of tangent of circle at (x 1 , y 1 ) , substitute xx 1 for x 2 , yy 1 for y 2, xy 1  x 1 y y  y1 x  x1 for x , for y and for xy and keep the constant as such. 2 2 2 E3 : Q Q 60 Tangen t xx 1  yy 1  g(x  x 1 )  f (y  y 1 )  c  0 Note Q P Q  This method of tangent at (x 1 , y 1 ) is applied any conics of second degree. i.e., equation of tangent of ID ax 2  2hxy  by 2  2 gx  2 fy  c  0 at (x 1 , y 1 ) is axx 1  h(xy 1  x 1 y )  byy 1  g(x  x 1 )  f (y  y 1 )  c  0 (2) Parametric form : Since parametric co-ordinates of circle x 2  y 2  a 2 is (a cos  , a sin  ), then U equation of tangent at (a cos  , a sin  ) is x. a cos   y. a sin   a 2 or x cos   y sin   a. (3) Slope form : Let y  mx  c is the tangent of the circle x 2  y 2  a 2. Length of perpendicular from centre of circle (0, 0) on line (y  mx  c) = radius of circle | c|  1m D YG  2  a  c a 1  m2 Substituting this value of c in y  mx  c, we get y  mx  a 1  m 2. Which are the required equations of tangents. Note : The reason why there are two equations y  mx  a 1  m 2 is that there are U two tangents, both are parallel and at the ends of a diameter.  The line ax  by  c  0 is a tangent to the circle x 2  y 2  r 2 if and only if ST c 2  r 2 (a 2  b 2 ).  The condition that the line lx  my  n  0 touches the x 2  y 2  2 gx  2 fy  c  0 is (lg  mf  n)2  (l 2  m 2 )(g 2  f 2  c).  Equation of tangent to the circle x 2  y 2  2 gx  2 fy  c  0 in terms of slope is y  mx  mg  f  (g 2  f 2  c) (1  m 2 ) (4) Point of contact : If circle be x 2  y 2  a 2 and tangent in terms of slope be y  mx  a (1  m 2 ) , circle Circle and System of Circles 91 Solving x 2  y 2  a 2 and y  mx  a (1  m 2 ) simultaneously, we get x   am (1  m 2 ) and a y (1  m 2 ) 60   am a  Thus, the co-ordinates of the points of contact are   ,  2 2  (1  m ) (1  m )   Alternative method : Let point of contact be (x 1 , y 1 ) then tangent at (x 1 , y 1 ) of x 2  y 2  a 2 is  x1   am (1  m 2 ) a and y1   x1 y  a2  1  m 1  a 1  m2 E3 xx 1  yy 1  a 2. Since xx 1  yy 1  a 2 and y  mx  a (1  m 2 ) are identical ,  (1  m 2 ) : If the line y = mx + c is the tangent to the circle x 2  y 2  r 2 then point of contact is   2 2 U Note ID   am a  Thus, the co-ordinates of the points of contact are   ,  2 2  (1  m ) (1  m )    given by   mr , r     c  c D YG  If the line ax+by+c = 0 is the tangent to the circle x2+y2=r2 then point of contact is given  2 2  by   ar ,  br     Example : 16 c c  The equations to the tangents to the circle x 2  y 2  6 x  4 y  12 which are parallel to the straight line 4x+3y+5=0, are [ISM Dhanbad 1973, MP PET 1991] (b) 4 x  3y  19  0, 4 x  3y  31  0 (c) 4 x  3y  19  0, 4 x  3y  31  0 (d) 3 x  4 y  19  0, 3 x  4 y  31  0 U (a) 3 x  4 y  19  0, 3 x  4 y  31  0 ST Solution : (c) Let equation of tangent be 4 x  3 y  k  0, then 9  4  12  4 (3)  3(2)  k 16  9  6  k   25  k  19 and – 31 Hence the equations of tangents are 4 x  3 y  19  0 and 4 x  3 y  31  0 Example : 17 The equations of any tangents to the circle x 2  y 2  2 x  4 y  4  0 is (a) y  m (x  1)  3 1  m 2  2 (b) y  mx  3 1  m 2 (c) y  mx  3 1  m 2  2 (d) None of these Solution : (a) Equation of circle is (x  1)  (y  2)  3. 2 2 2 As any tangent to x 2  y 2  3 2 is given by y  mx  3 1  m 2 Any tangent to the given circle will be y  2  m (x  1)  3 1  m 2  y  m (x  1)  3 1  m 2  2 92 Circle and System of Circles Example : 18 If a circle, whose centre is (–1, 1) touches the straight line x  2 y  12  0, then the coordinates of the point of contact are [MP PET 1998] 21   18 , (b)    5 5    7  (a)   ,  4  2   (c) (2, –7) (d) (– 2, – 5) Solution : (b) Let point of contact be P(x 1 , y1 ). This point lies on the given line ,  x 1  2y1  12 y1  1 , x1  1 P(x1,y1 ) 60 Gradient of OP  m 1  ….. (i) Gradient of x  2 y  12  m 2   O(– 1,1) E3 Both are perpendicular,  m1 m 2  1 1 2  y 1   1    1    1  y 1  1  2 x 1  2  2 x 1  y1  3   x1  1   2  ….. (ii)  18 21  , On solving the equation (i) and (ii), (x 1 , y 1 )    5   5 ID 4.8 Length of Tangent. From any point, say P(x 1 , y 1 ) two tangents can be drawn to a circle which are real, U coincident or imaginary according as P lies outside, on or inside the circle. Le PQ and PR be two tangents drawn from P(x 1 , y 1 ) to the circle tangent is called the length of D YG x 2  y 2  2 gx  2 fy  c  0. Then PQ =PR drawn from point P and is given by PQ = Q S1 R P (x1,y1) R PR  x 12  y 12  2 gx 1  2 fy1  c  S 1 4.9 Pair of Tangents. From a given point P(x 1 , y 1 ) two tangents PQ and PR can be Q drawn to the circle S  x 2  y 2  2 gx  2 fy  c  0. Their combined P (x1,y1) U equation is SS 1  T 2. Where S  0 is the equation of circle, T  0 is the equation of tangent at (x 1 , y 1 ) and S1 is obtained by replacing x by x1 and y by y1 ST in S. R 4.10 Power of Point with respect to a Circle. Let P(x 1 , y 1 ) be a point outside the circle and PAB and PCD drawn two secants. The power of P(x 1 , y 1 ) with respect to S  x 2  y 2  2 gx  2 fy  c  0 is equal to PA. PB which is x 12  y12  2 gx1  2 fy1  c  0  S 1  0  T B C D A Power remains constant for the circle i.e., independent of A and B.  PA. PB  PC. CD  (PT )2  S 1  ( S 1 )2  PA. PB  ( S 1 )2  square of the length of tangent. P(x1,y1 ) Circle and System of Circles 93 Note :  If P is outside, inside or on the circle then PA. PB is +ve, –ve or zero respectively. Important Tips The length of the tangent drawn from any point on the circle x  y  2 gx  2 fy  c  0 is 2  2 x 2  y 2  2 gx  2 fy  c1  0 c  c1. If two tangents drawn from the origin to the circle x 2  y 2  2 gx  2 fy  c  0 are perpendicular to each other, then g 2  f 2  2c. If the tangent to the circle x 2  y 2  r 2 at the point (a, b) meets the coordinate axes at the points A and B and O is the origin, then the area of the triangle OAB is r4. 2ab  If  is the angle subtended at P (x1 , y1 ) by the circle S  x 2  y 2  2 gx  2 fy  c  0, then cot  2 S1  ID  E3  to the circle 60  g  f2 c 2   a The angle between the tangents from (,  ) to the circle x 2  y 2  a 2 is 2 tan 1  2     2  a2   If OA and OB are the tangents from the origin to the circle x 2  y 2  2 gx  2 fy  c  0 and C is the centre of the circle, Example : 19 U c (g 2  f 2  c). D YG then the area of the quadrilateral OACB is  .   If the distances from the origin to the centres of three circles x 2  y 2  2i x  c 2  0 (i  1, 2, 3) are in G.P. then the lengths of the tangents drawn to them from any point on the circle x 2  y 2  c 2 are in (a) A.P. (b) G.P. (c) H.P. (d) None of these Solution : (b) The centres of the given circles are (i , 0) (i  1, 2, 3) The distances from the origin to the centres are i (i  1, 2, 3). It is given that  22  1  3. h2  k 2  c2 Let P(h, k ) be any point on the circle x 2  y 2  c 2 , then, U Now, Li = length of the tangent from (h, k) to x 2  y 2  2i x  c 2  0 ST  h 2  k 2  2ih  c 2 = c 2  2ih  c 2  2ih [ h 2  k 2  c 2 and i  1, 2, 3] [  22  1  3 ] Therefore, L22  22 h  2h ( 13 )  2h1 2h3  L1 L3. Hence, L1 , L2 , L3 are in G.P. Example : 20 From a point on the circle x 2  y 2  a 2 , two tangents are drawn to the circle x 2  y 2  a 2 sin 2 . The angle between them is [Rajasthan 2002] (a)  (b)  2 (c) 2  (d) None of these Solution : (c) Let any point on the circle x 2  y 2  a 2 be (a cos t, a sin t) and OPQ   Q Now; PQ = length of tangent from P on the circle x  y  a sin  2 2 2 2  O  PQ  a2 cos 2 t  a2 sin 2 t  a2 sin 2   a cos  R P PET 94 Circle and System of Circles OQ  Radius of the circle x 2  y 2  a 2 sin 2  OQ  a sin  ,  tan   OQ  tan      ;  PQ Angle between tangents QPR  2. Alternative Method : We know that, angle between the tangent from (,  ) to the circle x 2  y 2  a 2  . Let point on the circle x 2  y 2  a 2 be (a cos t, a sin t)    a sin  Angle between tangent = 2 tan 1   a 2 cos 2 t  a 2 sin 2 t  a 2 sin 2      2 tan 1  a sin    2  a cos       Two tangents to the circle x 2  y 2  4 at the points A and B meet at P (– 4, 0). The area of quadrilateral PAOB, where O is the origin, is (b) 6 2 Solution : (c) Clearly, sin   2 1  , 4 2 (c) 4 3    30 o. So area (POA )   Area (quadrilateral PAOB)  2. (d) None of these 1. 2. 4. sin 60 o 2 ID (a) 4 E3 Example : 21 60  a is 2 tan 1    2   2  a2  1 3. 2. 4 sin 60 o  8.  4 3. 2 2 P A 2  2 (–4,0) 2 O(0,0) Example : 22 The angle between a U Trick : Area of quadrilateral  r S 1  2 12  4 3 pair of tangents drawn from a point to the circle [IIT 1996] (a) x  y  4 x  6 y  4  0 (b) x 2  y 2  4 x  6 y  9  0 (c) x 2  y 2  4 x  6 y  4  0 (d) x 2  y 2  4 x  6 y  9  0 2 P 2 . The equation of the locus of the point P is D YG x 2  y 2  4 x  6 y  9 sin 2   13 cos 2   0 is B 2 Solution : (d) The centre of the circle x 2  y 2  4 x  6 y  9 sin 2   13 cos 2   0 is C (2, 3) and its radius is 2 2  (3)2  9 sin 2   13 cos 2   4  9  9 sin 2   13 cos 2   2 sin  Let P (h, k) be any point on the locus. The APC  . Also PAC   / 2 i.e. triangle APC is a right angle triangle. AC  PC U Thus sin   2 sin  A (h  2) 2  (k  3) 2 (h  2) 2  (k  3) 2  2  (h  2) 2  (k  3) 2  4 or h 2  k 2  4 h  6k  9  0 ST   C (–2,3) P(h,k) B Thus the required equation of the locus is x 2  y 2  4 x  6 y  9  0. 4.11 Normal to a Circle at a given Point. The normal of a circle at any point is a straight line, which is perpendicular to the tangent at the point and always passes through the centre of the circle. (1) Equation of normal: The equation of normal to the circle x 2  y 2  2 gx  2 fy  c  0 at any point (x 1 , y 1 ) is y  y 1  y1  f x  x 1 y  y1 (x  x 1 ) or  x1  g x 1  g y1  f P Normal 90° Tangen t Circle and System of Circles 95 Note :  The equation of normal to the circle x 2  y 2  a 2 at any point (x 1 , y 1 ) is xy 1  x 1 y  0 or x y  x 1 y1 60  The equation of any normal to the circle x 2  y 2  a 2 is y  mx where m is the slope of normal.  The equation of any normal to the circle x 2  y 2  2 gx  2 fy  c  0 is y  f  m (x  g). E3 where m is the slope of normal.  If the line y  mx  c is a normal to the circle with radius r and centre at (a, b) then b  ma  c. (2) Parametric form : Since parametric co-ordinates of circle x 2  y 2  a 2 is (a cos  , a sin  ). x y x y or   cos  sin  a cos  a sin  y  x tan  or y  mx where m  tan  , which is slope form of normal. Example : 23 The line lx  my  n  0 is a normal to the circle x 2  y 2  2 gx  2 fy  c  0, if U or ID  Equation of normal at (a cos  , a sin  ) is PET 1995] (a) lg mf  n  0 (c) lg mf  n  0 (b) lg  mf  n  0 [MP (d) lg mf  n  0 D YG Solution : (a) Since normal always passes through centre of circle, therefore (g,  f ) must lie on lx  my  n  0. Hence, lg mf  n  0 4.12 Chord of Contact of Tangents. (1) Chord of contact : The chord joining the points of contact of the two tangents to a conic drawn from a given point, outside it, is called the chord of contact (x,y P ) of tangents. U (2) Equation of chord of contact : The equation of the chord of contact of tangents drawn from a point (x 1 , y 1 ) to the circle x 2  y 2  a 2 is xx 1  yy 1  a 2. of chord ST Equation A (x1,y1 ) Chord of contact (x,y)Q of contact at (x 1 , y 1 ) to the circle x 2  y 2  2 gx  2 fy  c  0 is xx 1  yy 1  g (x  x 1 )  f (y  y1 )  c  0. It is clear from the above that the equation to the chord of contact coincides with the equation of the tangent, if point (x 1 , y 1 ) lies on the circle. The length of chord of contact  2 r 2  p 2 ; (p being length of perpendicular from centre to the chord) Area of APQ is given by a(x 12  y 12  a 2 )3 / 2. x 12  y 12 (3) Equation of the chord bisected at a given point : The equation of the chord of the circle S  x 2  y 2  2 gx  2 fy  c  0 bisected at the point (x 1 , y 1 ) is given by T  S ' 96 Circle and System of Circles i.e. xx 1  yy 1  g (x  x 1 )  f (y  y 1 )  c  x 12  y 12  2 gx 1  2 fy1  c. Example : 24 Tangents are drawn from any point on the circle x 2  y 2  a 2 to the circle x 2  y 2  b 2. If the chord of contact touches the circle x 2  y 2  c 2 , a  b, then [MP PET 1999, Rajasthan PET 1999] ax cos   ay sin   b 2 This chord touches the circle x 2  y 2  c 2 , c Example : 25 b2 a cos 2   sin 2  E3 Hence, Radius = Perpendicular distance of chord from centre. 60 (a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) a, c, b are in G.P. Solution : (b) Chord of contact of any point (a cos  , a sin ) on 1st circle with respect to 2nd circle is  b 2  ac.Hence a,b,c are in G.P. The area of the triangle formed by the tangents from the point (4, 3) to the circle x 2  y 2  9 and the [IIT 1981, MP PET 1991] (a) 25 sq. units 192 (b) 192 sq. units 25 ID line joining their points of contact is 384 sq. units 25 (c) (d) None of these equation of OP is y  3 x. 4 U Solution : (b) The equation of the chord of contact of tangents drawn from P (4, 3) to x 2  y 2  9 is 4 x  3 y  9. The Q D YG 9 Now, OM = (length of the perpendicular from (0, 0) on 4 x  3 y  9  0)  5 P  QR  2. QM  2 OQ 2  OM 2  2 9  Now, PM  OP  OM  5  Example : 26 (4,3 ) 81 24  25 5 M R (0, O0) x2+y2=9 1  24   16  192 9 16 sq. units .So, Area of  PQR      2  5  5  25 5 5 The locus of the middle points of those chords of the circle x 2  y 2  4 which subtend a right angle at the origin is [MP PET 1990; IIT 1984; Rajasthan PET 1997; DCE 2000, 01] U (a) x 2  y 2  2 x  2y  0 (b) x 2  y 2  4 (c) x 2  y 2  2 (d) (x  1)2  (y  2)2  5 Solution : (c) Let the mid-point of chord is (h, k). Also radius of circle is 2. Therefore ST OC  cos 45 o  OB Y h2  k 2 1   h2  k 2  2 2 2 B C (h,k) 45 ° O Hence locus is x 2  y 2  2 Example : 27 A X If two distinct chords, drawn from the point (p, q) on the circle x 2  y 2  px  qy (where p, q  0) are bisected x-axis, then by the [IIT 1999] (a) p 2  q 2 (b) p 2  8q 2 (c) p 2  8q 2 (d) p 2  8q 2 Solution : (d) Let (h, 0) be a point on x-axis, then the equation of chord whose mid-point is (h, 0) will be 1 1 1 1 xh  p (x  h)  q (y  0)  h 2  ph. This passes through (p, q), hence ph  p ( p  h)  q.q  h 2  ph 2 2 2 2 Circle and System of Circles 97  ph  1 2 1 3 1 1 p  ph  q 2  h 2  ph  h 2  ph  ( p 2  q 2 )  0 ;  2 2 2 2 2 h is real, hence B 2  4 AC  0 9 2 1 p  4. (p 2  q 2 )  0  9 p 2  8 (p 2  q 2 )  0  p 2  8 q 2  0 4 2   p 2  8q 2 4.13 Director Circle. 60 The locus of the point of intersection of two perpendicular tangents to a circle is called the Director circle. Let the circle be x 2  y 2  a 2 , then equation of the pair of (x  y  a 2 2 to 2 ) (x 12 a  y 12 circle from a point is P(x1,y1)  a )  (xx 1  yy 1  a ). If this represents a pair of 2 2 2 E3 tangents 90 ° perpendicular lines, coefficient of x 2  coefficient of y 2  0 i.e. (x 12  y12  a 2  x 12 )  (x 12  y12  a 2  y12 )  0  x 12  y 12  2a 2 ID Hence the equation of director circle is x 2  y 2  2a 2. Obviously director circle is a concentric circle whose radius is given circle.  : Director x 2  y 2  2 gx  2 fy  2c  g 2  f 2  0. of circle x 2  y 2  2 gx  2 fy  c  0 D YG 4.14 Diameter of a Circle. circle is U Note 2 times the radius of the The locus of the middle points of a system of parallel chords of a circle is called a diameter of the circle. The equation of the diameter bisecting parallel chords Diamete y  mx  c (c is a parameter) of the circle x 2  y 2  a 2 is x  my  0. Note : The diameter corresponding to a system of O P(h,k )y=mx+c A U parallel chords of a circle always passes through the centre of the circle and is perpendicular to the parallel chords. x+my=0 r B A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of the circle has the equation 2 x  y  2. Then the equation of the circle is ST Example : 28 (a) x 2  y 2  2 x  1  0 Solution : (b) (b) x 2  y 2  2 x  1  0 (c) x 2  y 2  2y  1  0 (d) None of these. The line joining (4, 3) and (2, 1) is also along a diameter. So, the centre is the intersection of the diameters 2 x  y  2 and y  3  (x  4 ). Solving these, the centre = (1, 0)  Radius = Distance between (1, 0) and (2, 1) = 2.  Equation of circle (x  1) 2  y 2  ( 2 ) 2  x 2  y 2  2 x  1  0 Example : 29 The diameter of the circle x 2  y 2  4 x  2y  11  0 corresponding to a system of chords parallel to the line x  2 y  1  0 (a) x  2 y  3  0 Solution : (c) (b) 2 x  y  3  0 (c) 2 x  y  3  0 (d) None of these The centre of the given circle is (2, –1) the equation of the line perpendicular to chord x  2 y  1  0 is 2 x  y  k  0 98 Circle and System of Circles Since the line passes through the point (2, –1) therefore k = – 3. The equation of diameter is 2 x  y  3  0. 4.15 Pole and Polar. Let P (x 1 , y 1 ) be any point inside or outside the circle. Draw chords AB and A' 60 B' passing through P. If tangents to the circle at A and B meet at Q (h, k), then locus of Q is called the polar of P with respect to circle and P is called the pole and if tangents to the circle at A' and B' meet at Q', then the straight line QQ' is polar with P as its pole. If circle be x 2  y 2  a 2 then AB is the chord of contact of Q (h, k), hx  ky  a 2 is its Pole P(x1,y1 ) Pola r A Q(h,k ) B Q Q(h,k ) A A Polar A B P(x1,y1 Pole ) ID Q E3 equation. But P (x 1 , y 1 ) lies on AB,  hx 1  ky1  a 2. B B U Hence, locus of Q (h, k) is xx 1  yy 1  a 2 , which is polar of P (x 1 , y 1 ) with respect to the circle x 2  y 2  a2. D YG (1) Coordinates of pole of a line : The pole of the line lx  my  n  0 with respect to the circle x 2  y 2  a 2. Let pole be ( x 1 , y 1 ), then equation of polar with respect to the circle x 2  y 2  a 2 is xx 1  yy 1  a 2  0 , which is same as lx  my  n  0  a 2l a 2m  x 1 y1 a2 a 2m a 2l .   ,  x1   and y 1  . Hence, the required pole is   , l m n n n n   n (2) Properties of pole and polar (i) If the polar of P (x 1 , y 1 ) w.r.t. a circle passes through Q (x 2 , y 2 ) then the polar of Q will Then U pass through P and such points are said to be conjugate points. (ii) If the pole of the line ax  by  c  0 w.r.t. a circle lies on another line a1 x  b1 y  c1  0; ST then the pole of the second line will lie on the first and such lines are said to be conjugate lines. (iii) The distance of any two points P (x 1 , y 1 ) and Q (x 2 , y 2 ) from the centre of a circle is proportional to the distance of each from the polar of the other. (iv) If O be the centre of a circle and P any point, then OP is perpendicular to the polar of P. (v) If O be the centre of a circle and P any point, then if OP (produced, if necessary) meet the polar of P in Q, then OP. OQ = (radius)2. Note :  Equation of polar is like as equation of tangent i.e., T = 0 (but point different)  Equation of polar of the circle x 2  y 2  2 gx  2 fy  c  0 with respect to (x 1 , y 1 ) is xx 1  yy 1  g(x  x 1 )  f (y  y 1 )  c  0  If the point P is outside the circle then equation of polar and chord of contact will coincide. In this case the polar cuts the circle at two points. Circle and System of Circles 99  If the point P is on the circle then equation of polar, chord of contact and tangent at P will coincide. So in this case the polar touches the circle.  If the point P is inside the circle (not its centre) then only its polar will exist. In this case the polar is outside the circle. The polar of the centre lies at infinity.  If a triangle is like that its each vertex is a pole of opposite side with respect to a circle then it is called self conjugate triangle. 60 1  Example : 30 The polar of the point  5 ,   with respect to circle (x  2) 2  y 2  4 is 2  1996] (a) 5 x  10 y  2  0 PET (d) x  10 y  2  0  5x  E3 1  The polar of the point  5 ,   is xx 1  yy 1  g(x  x 1 )  f (y  y1 )  c  0 2  Solution : (b) y 1 y  2 (x  5)  0  0  0  3 x   10  0  6 x  y  20  0. 2 2 The pole of the straight line 9 x  y  28  0 with respect to circle 2 x 2  2y 2  3 x  5 y  7  0 is MNR 1984; UPSEAT 2000] (a) (3, 1) (c) (3, – 1) 3 5 7 Equation of given circle is x 2  y 2  x  y   0 2 2 2 Solution : (c) (b) (1, 3) ID Example : 31 (c) 10 x  y  10  0 (b) 6 x  y  20  0 [Rajasthan 2 2 2 [Rajasthan PET 1990, 99; (d) (– 3, 1) 2 45 45 5 3 0  0 and the line 9 X  Y  and Y  y  , we get the equation of circle X 2  Y 2  4 2 8 4 D YG Put X  x  U 3  5 9 25 7 3 5 45      x    y       0   x    y    0 4 4 8 4  4  16 16 2    45 45  1 9  8 8 Hence pole   , 45  45 2  2   9 1    , . 4 4   But, x  1 5 9 3   3 and y     1 , hence the pole is 4 4 4 4 (3, – 1). 4.16 Two Circles touching each other. U (1) When two circles touch each other externally : Then distance between their centres = Sum of their radii i.e., | C1 C 2 |  r1  r2 ST In such cases, the point of contact P divides the line joining C1 C P r and C2 internally in the ratio r1 : r2  1  1 C 2 P r2 r1 C1 P r2 C2 If C1  (x 1 , y 1 ) and C 2  (x 2 , y 2 ) , then co-ordinate of P is  r1 x 2  r2 x 1 r1 y 2  r2 y1    , r1  r2   r1  r2 (2) When two circles touch each other internally : Then distance between their centres = Difference of their radii i.e., | C1 C 2 |  r1  r2 In such cases, the point of contact P divides the line joining C1 C P r and C2 externally in the ratio r1 : r2  1  1 C 2 P r2 If C1  (x 1 , y 1 ) and C 2  (x 2 , y 2 ) , then co-ordinate of P is C1 r2 C2 P r1 100 Circle and System of Circles  r1 x 2  r2 x 1 r1 y 2  r2 y1  , r1  r2  r1  r2    4.17 Common Tangents to Two circles. Different cases of intersection of two circles : Let the two circles be (x  x 1 )2  (y  y 1 )2  r12 60 …..(i) and (x  x 2 )2  (y  y 2 )2  r22 …..(ii) with centres C1 (x 1 , y 1 ) and C 2 (x 2 , y 2 ) and radii r1 and r2 respectively. Then following cases may arise : E3 Case I : When | C1 C 2 |  r1  r2 i.e., the distance between the centres is greater than the sum of radii. ID In this case four common tangents can be drawn to the two circles, in which two are direct common tangents and the other two are transverse common tangents. Direct common tangents r1 C1 T r2 C2 D Transvers common tangents U Case II : When | C 1 C 2 |  r1  r2 i.e., the distance between the centres is equal to the sum of radii. D YG In this case two direct common tangents are real and distinct while the transverse tangents are coincident. Case III : When ST In this case two direct common tangents are real and distinct while the transverse tangents are imaginary. Case IV : When C1 T C2 D Transverse common tangents | C1 C 2 |  r1  r2 i.e., the distance between the centres is less than sum of U radii. Direct common tangents Direct common tangents C2 C1 D | C1 C 2 |  | r1  r2 | , i.e., the distance between the centres is equal to the difference of the radii. In this case two tangents are real and coincident while the other two tangents are imaginary. C1 Tangent at the r2 point of C2 P contact r1 Circle and System of Circles 101 | C1 C 2 |  | r1  r2 | , i.e., the distance between the centres is less than the Case V : When difference of the radii. In this case, all the four common tangents are imaginary. Note 60 C1 r2 C2 r1 :  Points of intersection of common tangents : The points T1 and T 2 (points of E3 intersection of indirect and direct common tangents) divides C1 C 2 internally and externally in the ratio r1 : r2 respectively.  Equation of the common tangents at point of contact is S 1  S 2  0. 1  1 f 2  1 c2. ST U D YG U g 2 ID  If the circle x 2  y 2  2 gx  c 2  0 and x 2  y 2  2 fy  c 2  0 touch each other, then Circle and System of Circles 99 (i) Position Do not intersect or one outside the other C1 C 2  r1  r2 T1 C2 C1 C2 (iii) C1 C 2  r1  r2 One inside the other 0 C1 E3 C1 C 2  | r1  r2 | 4 T2 T1 External touch C2 C1 ID (ii) No. of common tangents Diagram 60 Condition 3 T2 C2 (v) C1 C 2  | r1  r2 | U Internal touch D YG (iv) Intersection at two real points | r1  r2 |  C1 C 2  r1  r2 Example : 32 C2 C1 2 T2 If circles x 2  y 2  2ax  c  0 and x 2  y 2  2by  c  0 touch each other, then (a) 1 1 1   a b c C 1  (a, 0), (b) 1 a2  1 b2  r1  a 2  c ; C 2  (0,  b), U Solution : (d) 1 C1 1 c2 (c) 1 1   c2 a b [MNR 1987] (d) 1 a2  1 b2  1 c r2  b 2  c ; C 1 C 2  a 2  b 2  Circles touch each other, therefore r1  r2  C1 C 2 a2  c  b 2  c  a2  b 2 ST  Multiplying by Example : 33 Solution : (a) 1 a2b 2c 2 , we get 1 a2  a2b 2  b 2c  a2c  0  1 b2  1. c If two circles (x  1)2  (y  3)2  r 2 and x 2  y 2  8 x  2y  8  0 intersect in two distinct points, then [IIT 1989; Karnataka CET 2002; DCE 2000, 01; AIEEE 2003] (a) 2  r  8 (b) r  2 (c) r  2 (d) r  2 When two circles intersect each other, then difference between their radii < Distance between their centres  r3  5  r  8 …..(i) Sum of their radii > Distance between their centres  r3 5  r  2 Hence by (i) and (ii), 2 < r < 8. …..(ii) 100 Circle and System of Circles Example : 34 The equation of the circle having the lines x 2  2 xy  3 x  6 y  0 as its normals and having size just sufficient to contain the circle x (x  4)  y (y  3)  0 is Solution : (b) [Roorkee 1990] (a) x  y  3 x  6 y  40  0 (b) x  y  6 x  3 y  45  0 (c) x  y  8 x  4 y  20  0 (d) x 2  y 2  4 x  8 y  20  0 2 2 2 2 2 2 Given pair of normals is x 2  2 xy  3 x  6 y  0 or (x  2y)(x  3)  0 60  Normals are x  2 y  0 and x  3  0 The point of intersection of normals x  2 y  0 and x  3  0 is the centre of C1 required circle, we get centre C1  (3, 3 / 2) and other circle is x (x  4)  y (y  3)  0 or x 2  y 2  4 x  3 y  0 C2 …..(i) 9 5  4 2 Since the required circle just contains the given circle (i), the given circle should touch the required circle internally fr om inside. E3 Its centre C2  (2, 3 / 2) and radius r  4  2 5 5 15 3 3 (3  2)2      5   2 2 2 2 2   Therefore, radius of the required circle | C1  C 2|  r  2 Example : 35 or x 2  y 2  6 x  3 y  45  0. The equation of the circle which touches the circle x 2  y 2  6 x  6 y  17  0 externally and to which the lines (a) x  y  6 x  2y  1  0 (c) x  y  6 x  6y  1  0 2 2 2 2  D YG Joint equations of normals are x 2  3 xy  3 x  9 y  0  [Roorkee 1994] (b) x  y  6 x  2y  1  0 (d) x 2  y 2  6 x  2y  1  0 U x 2  3 xy  3 x  9 y  0 are normals, is Solution : (d) 2 ID 3   15  Hence, equation of required circle is (x  3) 2   y      2    2  2 2 x (x  3y)  3 (x  3y)  0  (x  3) (x  3y)  0 Given normals are x  3  0 and x  3 y  0 , which intersect at centre of circle whose coordinates are (3, 1). The given circle is C1  (3,  3), r1  1 ; C 2  (3, 1), r2  ? If the two circles touch externally, then C1 C 2  r1  r2  4  1  r2  r2  3  Equation of required circle is (x  3)2  (y  1)2  (3)2  x 2  y 2  6 x  2y  1  0 Example : 36 The number of common tangents to the circles x 2  y 2  4 and x 2  y 2  6 x  8 y  24 is (a) 0 (b) 1 (c) 3 (d) 4 Solution : (b) Circles S 1  x 2  y 2  (2) 2 and S 2  (x  3) 2  (y  4 ) 2  (7) 2 [IIT 1998] U  Centres C1  (0, 0), C 2  (3, 4) and radii r1  2, r2  7 C 1 C 2  (3) 2  (4 ) 2  5 , r2  r1  7  2  5  C1 C 2  r2  r1 i.e. circles touch internally. Hence there is only one common tangent. There are two circles whose equations are x 2  y 2  9 and x 2  y 2  8 x  6 y  n 2  0, n  Z. If the two circles have exactly ST Example : 37 two common tangents, then the number of possible values of n is (a) 2 (b) 8 (c) 9 Solution : (c) (d) None of these For x  y  9 , the centre = (0, 0) and the radius = 3 2 2 For x 2  y 2  8 x  6 y  n 2  0. The centre = (4, 3) and the radius  (4 ) 2  (3) 2  n 2  4 2  3 2  n 2  0 or n 2  5 2 or 5  n  5. Circles should cut to have exactly two common tangents. So, r1  r2  C1 C 2 ,  3  25  n 2  (4 ) 2  (3) 2 or 25  n 2  2 or 25  n 2  4  n 2  21 or  21  n  21 Therefore, common values of n should satisfy  21  n  21. But n Z , So, n  4,  3,........ 3, 4.  Number of possible values of n = 9. Circle and System of Circles 101 4.18 Common chord of two Circles. (1) Definition : The chord joining the points of intersection of two given circles is called their common chord. (2) Equation of common chord : The equation of the common chord of two circles ….(i) S 2  x 2  y 2  2 g 2 x  2 f2 y  c 2  0 ….(ii) M P C1 C2 60 and S 1  x 2  y 2  2 g1 x  2 f1 y  c1  0 is 2 x (g1  g 2 )  2 y ( f1  f2 )  c1  c 2  0 i.e. S 1  S 2  0. S1=0 S2=0 Q (3) Length of the common chord : PQ  2 (PM )  2 C1 P 2  C1 M 2 Note E3 Where C 1 P  radius of the circle S  0 and C 1 M  length of the perpendicular from the centre C 1 to the common chord PQ. The length of the common chord is 2 r12  p12  2 r22  p 22 where p1 and p2 :  ID are the lengths of perpendicular drawn from the centre to the chord. If the common chord of the circles x 2  (y  ) 2  16 and x 2  y 2  16 subtend a right angle at the origin, then  is equal to D YG Example : 38 U  While using the above equation of common chord the coefficient of x 2 and y 2 in both equation should be equal.  Two circle touches each other if the length of their common chord is zero.  Maximum length of the common chord = diameter of the smaller circle. (a) 4 Solution : (c) (b) (c) 4 2 4 2 (d) 8 The common chord of given circles is S 1  S 2  0  x 2  (y   ) 2  16  { x 2  y 2  16 }  0 i.e., y   (    0) 2 The pair of straight lines joining the origin to the points of intersection of y   2y  and x 2  y 2  16 is x 2  y 2  16   2    Example : 39 Which of the following is a point on the common chord of the circles x 2  y 2  2 x  3y  6  0 x  y  x  8 y  13  0 2 ST 2 (a) (1, –2) Solution : (d) 2  2 x 2  ( 2  64 ) y 2  0. These lines are at right angles if 2  2  64  0 , i.e.,    4 2. U   [Karnataka CET 2003] (b) (1, 4) Given circles are, S 1  x  y  2 x  3 y  6  0 2 and 2 (d) (1, – 4) (c) (1, 2) ….. (i) and S 2  x  y  x  8 y  13  0 2 2 ….. (ii)  Equation of common chord is S1  S 2  0  x  5 y  19  0 , and out of the four given points only point (1, – 4) satisfies it. Example : 40 If the circle x 2  y 2  4 bisects the circumference of the circle x 2  y 2  2 x  6 y  a  0, then a equals [Rajasthan PET 1999] (a) 4 Solution : (c) (b) – 4 The common chord of given circles is S 1  S 2  0  (c) 16 2 x  6y  4  a  0 (d) – 16 …..(i) Since, x 2  y 2  4 bisects the circumferences of the circle x 2  y 2  2 x  6 y  a  0, therefore (i) passes through the centre of second circle i.e. (1, – 3).  2 + 18 – 4 – a = 0  a = 16. 102 Circle and System of Circles 4.19 Angle of Intersection of Two Circles. The angle of intersection between two circles S = 0 and S' = 0 is defined as the angle between their tangents at their point of intersection. A B If S  x 2  y 2  2 g1 x  2 f1 y  c1  0 S=0 S=0 – r1 S '  x 2  y 2  2 g 2 x  2 f2 y  c 2  0 P  r2 C2 60 C1 B A Q are two circles with radii r1 , r2 and d be the distance between their centres then the angle of intersection  between r12  r12  d 2 2 (g1 g 2  f1 f2 )  (c1  c 2 ) or cos   2r1 r2 2 g12  f12  c1 g 22  f22  c 2 E3 them is given by cos   Note ID (1) Condition of Orthogonality : If the angle of intersection of the two circles is a right angle (  90 o ) , then such circles are called orthogonal circles and condition for their orthogonality is P 2 g1 g 2  2 f1 f2  c1  c 2 : 90° C1 (–g1,–f1) When the two circles intersect orthogonally then the length C2 (–g2,–f2) U of tangent on one circle from the centre of other circle is equal to the radius of the other circle.  Equation of a circle intersecting the three circles y f1 f2 f3 1 1 0 1 1 A circle passes through the origin and has its centre on y  x. If it cuts x 2  y 2  4 x  6 y  10  0 orthogonally, then the equation of the circle is [EAMCET 1994] (a) x2  y2  x  y  0 (b) x 2  y 2  6x  4y  0 (c) x 2  y 2  2 x  2y  0 Let the required circle be x 2  y 2  2 gx  2 fy  c  0 x 2  y 2  2 x  2y  0 (d).......(i) U Solution : (c) x g1 g2 g3 D YG orthogonally is Example : 41 x2  y2  c1  c2  c3 x 2  y 2  2 g i x  2 fi y  ci  0 (i  1, 2, 3) ST This passes through (0, 0), therefore c = 0 The centre (g,  f ) of (i) lies on y = x, hence g = f. Since (i) cuts the circle x 2  y 2  4 x  6 y  10  0 orthogonally, therefore 2 (2 g  3 f )  c  10   10 g  10  g  f  1 Example : 42 The centre of the circle, ( g  f and c  0 ). Hence the required circle is x 2  y 2  2 x  2y  0. which cuts orthogonally each of the three circles x 2  y 2  2 x  17 y  4  0, x 2  y 2  7 x  6 y  11  0 and x 2  y 2  x  22 y  3  0 is (a) (3, 2) Solution : (a) (b) (1, 2) Let the circle is x  y  2 gx  2 fy  c  0 2 2 (c) (2, 3) [MP PET 2003] (d) (0, 2) …..(i) Circle (i) cuts orthogonally each of the given three circles. Then according to condition 2g1 g2  2 f1 f2  c1  c2 2 g  17 f  c  4 …..(ii) 7 g  6 f  c  11 …..(iii)  g  22 f  c  3 …..(iv) On solving (ii), (iii) and (iv), g  3, f  2. Therefore, the centre of the circle (g,  f )  (3, 2) Circle and System of Circles 103 Example : 43 The locus of the centre of a circle which cuts orthogonally the circle x 2  y 2  20 x  4  0 and which touches x  2 is [UPSEAT 2001] (a) Solution : (d) y  16 x  4 2 (b) x 2  16 y (c) x  16 y  4 2 (d) y  16 x 2 Let the circle be x  y  2 gx  2 fy  c  0 2 …..(i) 2 It cuts the circle x 2  y 2  20 x  4  0 orthogonally 2 (10 g  0  f )  c  4   20 g  c  4 …..(ii) Circle (i) touches the line x  2;  x  0 y  2  0  g02  g 2  f 2  c  (g  2) 2  g 2  f 2  c  4g  4  f 2  c 1 4.20 Family of Circles. …..(iii) E3 Eliminating c from (ii) and (iii), we get  16 g  4  f 2  4  f 2  16 g  0. Hence the locus of ( g,  f ) is y 2  16 x  0  y 2  16 x. 60  (where  is a parameter,   1) U S  S ' 0 ID (1) The equation of the family of circles passing through the point of intersection of two given circles S = 0 and S' = 0 is given as S=0 S=0 S+S=0 D YG (2) The equation of the family of circles passing through the point of intersection of circle S = 0 and a line L = 0 is given as S  L  0 (where  is a parameter) S=0 S+L=0 L=0 (3) The equation of the family of circles touching the circle S = 0 and the line L = 0 at their point of contact P is (where  is a parameter) ST U S  L  0 S=0 L=0 S+L=0 (4) The equation of a family of circles passing through two given points P (x 1 , y 1 ) and Q (x 2 , y 2 ) can be written in the form x  (x  x 1 )(x  x 2 )  (y  y 1 )(y  y 2 )   x 1 x2 y y1 y2 1 1 0 1 P(x1,y1) Q(x2,y2) (where  is a parameter) (5) The equation of family of circles, which touch y  y 1  m (x  x 1 ) at (x 1 , y 1 ) for any finite m is (x  x 1 )2  (y  y1 )2   {(y  y1 )  m (x  x 1 )}  0 (x1,y1) y–y1=m(x–x1) 104 Circle and System of Circles And if m is infinite, the family of circles is (x  x 1 )2  (y  y1 )2   (x  x 1 )  0 (where  is a parameter) (6) Equation of the circles given in diagram is (x  x 1 )(x  x 2 )  (y  y 1 )(y  y 2 )  cot  {( x  x 1 )(y  y 2 ) 60  (x  x 2 )(y  y 1 )}  0  (x1,y1) E3 (x2,y2)  The equation of the circle through the points of intersection of x 2  y 2  1  0, x 2  y 2  2 x  4 y  1  0 and touching the line x  2 y  0 is x  y  x  2y  0 2 (a) (b) x  y  x  20  0 2 2 (c) [Roorkee 1989] x  y  x  2y  0 2 2 (d) 2 (x  y )  x  2 y  0 2 2 Family of circles is x  y  2 x  4 y  1   (x  y  1)  0 2 x2  y2  2 2 2 4 1 x y 0 1 1 1 2 U Solution : (c) 2 ID Example : 44 2 2 D YG 2   2  1     1   1 , Centre is   and radius   1      1      1       1  1          4  2 (1   ) 2 Since it touches the line x  2 y  0, Hence Radius = perpendicular distance from centre to the line   4  2 1   2  4  2  1 1 4  1 1 12  2 2 5  4  2     1   1 cannot be possible in case of circle, so   1.  Equation of circle is x 2  y 2  x  2y  0 Example : 45 The equation of the circle through the points of intersection of the circles x 2  y 2  6 x  2 y  4  0, U x 2  y 2  2 x  4 y  6  0 and with its centre on the line y  x (a) (b) 7 x 2  7 y 2  10 x  10 y  12  0 (c) 7 x 2  7 y 2  10 x  10 y  12  0 (d) 7 x 2  7 y 2  10 x  10 y  12  0 Equation of any circle through the points of intersection of given circles is (x 2  y 2  6 x  2 y  4 )   (x 2  y 2  2 x  4 y  6)  0  x 2 (1   )  y 2 (1   )  2 x (3   )  2 y (1  2 )  (4  6  )  0 ST Solution : (b) 7 x 2  7 y 2  10 x  10 y  12  0 or, x 2  y 2  2 x (3  ) 2y (1  2) (4  6 )   0 (1  ) (1  ) (1  ) …..(i) 2  1 3    3   2  1  ,  Its centre    2  1  3      1   lies on the line y = x. Then 1 1 1   1    4  3  4    3 4 Substituting the value of   in (i), we get the required equation as 7 x 2  7 y 2  10 x  10 y  12  0. 3 4.21 Radical Axis. The radical axis of two circles is the locus of a point which moves such that the lengths of the tangents drawn from it to the two circles are equal. Circle and System of Circles 105 Consider, S  x 2  y 2  2 gx  2 fy  c  0 …..(i) and S '  x 2  y 2  2 g1 x  2 f1 y  c1  0 …..(ii) Let P (x 1 , y 1 ) be a point such that | PA |  | PB |  (x 12  y 12  2 gx 1  2 fy1  c)  (x 12  y 12  2 g1 x 1  2 f1 y 1  c1 ) On squaring, x 12  y 12  2 gx1  2 fy1  c  x 12  y 12  2 g1 x 1  2 f1 y 1  c1 60  2 (g  g1 ) x 1  2 ( f  f1 ) y 1  c  c1  0  Locus of P (x 1 , y 1 ) is 2 (g  g 1 ) x  2 ( f  f1 ) y  c  c1  0 P(x1,y1) A E3 P(x1,y1) B C1 C2 C1 S=0 C2 ID S=0 B A D YG U which is the required equation of radical axis of the given circles. Clearly this is a straight line. (1) Some properties of the radical axis (i) The radical axis and common chord are identical : Since the radical axis and common chord of two circles S = 0 and S' = 0 are the same straight line S – S' = 0, they are identical. The only difference is that the common chord exists only if the circles intersect in two real points, while the radical axis exists for all pair of circles irrespective of their position (Except when one circle is inside the other). C1 Common chord Common tangent Radical axis C1 C2 C1 C2 Intersecting circles Touching circles Non intersecting circles C2 (ii) The radical axis is perpendicular to the straight line which joins the centres of the circles : Consider, S  x 2  y 2  2 gx  2 fy  c  0 U and …..(i) S 1  x 2  y 2  2 g1 x  2 f1 y  c1  0 …..(ii) ST Since C1  ( g,  f ) and C 2  ( g1 ,  f1 ) are the centres of the circles P(x1,y1) f  f1  f1  f (i) and (ii), then slope of the straight line C1 C 2    m 1 (say)  g1  g g  g 1 Equation of the radical axis is, 2 (g  g1 ) x  2 ( f  f1 ) y  c  c1  0 Slope of radical axis is  (g  g1 )  m 2 (say).  m 1 m 2  1 ( f  f1 ) B A R C1 C2 Q S=0 S=0 Hence C1 C 2 and radical axis are perpendicular to each other. (iii) The radical axis bisects common tangents of two circles : Let AB be the common tangent. If it meets the radical axis LM in M then MA and MB are two tangents to the circles. Hence MA = MB, since length of tangents are equal from any point on radical axis. Hence radical axis bisects the common tangent AB. A M C1 L B C2 T T C1 A B C2 C1 C2 A 106 Circle and System of Circles 60 If the two circles touch each other externally or internally then A and B coincide. In this case the common tangent itself becomes the radical axis. (iv) The radical axis of three circles taken in pairs are concurrent : Let the equations of three circles be S 1  x 2  y 2  2 g1 x  2 f1 y  c1  0 S 3  x 2  y 2  2 g 3 x  2 f3 y  c 3  0 E3 S 2  x 2  y 2  2 g 2 x  2 f2 y  c 2  0 …..(i) …..(ii) …..(iii) The radical axis of the above three circles taken in pairs are given by …..(iv) S 2  S 3  2 x (g 2  g 3 )  2 y ( f2  f3 )  c 2  c 3  0 …..(v) S 3  S 1  2 x (g 3  g1 )  2 y ( f3  f1 )  c 3  c1  0.....(vi) U ID S 1  S 2  2 x (g1  g 2 )  2 y ( f1  f2 )  c1  c 2  0 Adding (iv), (v) and (vi), we find L.H.S. vanished identically. Thus the three lines are concurrent. D YG (v) If two circles cut a third circle orthogonally, the radical axis of the two circles will pass through the centre of the third circle or The locus of the centre of a circle cutting two given circles orthogonally is the radical axis of the two circles. Let S 1  x 2  y 2  2 g1 x  2 f1 y  c1  0 …..(i) S 2  x 2  y 2  2 g 2 x  2 f2 y  c 2  0 …..(ii) S 3  x 2  y 2  2 g 3 x  2 f3 y  c 3  0 …..(iii) Since (i) and (ii) both cut (iii) orthogonally,  2 g1 g 3  2 f1 f3  c1  c 3 and 2 g 2 g 3  2 f2 f3  c 2  c 3 U Subtracting, we get 2 g 3 (g1  g 2 )  2 f3 ( f1  f2 )  c1  c 2 …..(iv) Now radical axis of (i) and (ii) is S 1  S 2  0 or 2 x (g1  g 2 )  2 y ( f1  f2 )  c1  c 2  0 ST Since it will pass through the centre of circle (iii)   2 g 3 (g1  g 2 )  2 f3 ( f1  f2 )  c1  c 2  0 or 2 g 3 (g1  g 2 )  2 f3 ( f1  f2 )  c1  c 2 …..(v) which is true by (iv). Note : Radical axis need not always pass through the mid point of the line joining the centres of the two circles. 4.22 Radical Centre. The radical axes of three circles, taken in pairs, meet in a point, which is called their radical centre. Let the three circles be S 1  0.....(i) , S 2  0.....(ii) , S 3  0.….(iii) L S1=0 Let OL, OM and ON be radical axes of the pair sets of circles S2=0 O M N S3=0 Circle and System of Circles 107 {S 1  0, S 2  0}, {S 3  0, S 1  0} and {S 2  0, S 3  0} respectively. Equation of OL, OM and ON are respectively S 1  S 2  0.....(iv) , S 3  S 1  0.....(v), S 2  S 3  0.....(vi) 60 Let the straight lines (iv) and (v) i.e., OL and OM meet in O. The equation of any straight line passing through O is (S 1  S 2 )   (S 3  S 1 )  0 where  is any constant For   1 , this equation become S 2  S 3  0 , which is by (vi), equation of ON. E3 Thus the third radical axis also passes through the point where the straight lines (iv) and (v) meet. In the above figure O is the radical centre. (1) Properties of radical centre (i) Co-ordinates of radical centre can be found by solving the equations A F S1  S 2  S 3  0 ID (ii) The radical centre of three circles described on the sides of a triangle as diameters is the orthocentre of the triangle : Draw perpendicular from A on BC.  ADB  ADC   / 2 B E I C D D YG U Therefore, the circles whose diameters are AB and AC passes through D and A. Hence AD is their radical axis. Similarly the radical axis of the circles on AB and BC as diameter is the perpendicular line from B on CA and radical axis of the circles on BC and CA as diameter is the perpendicular line from C on AB. Hence the radical axis of three circles meet in a point. This point I is radical centre but here radical centre is the point of intersection of altitudes i.e., AD, BE and CF. Hence radical centre = orthocentre. (iii) The radical centre of three given circles will be the centre of a fourth circle which cuts all the three circles orthogonally and the radius of the fourth circle is the length of tangent drawn from radical centre of the three given circles to any of these circles. Let the fourth circle be (x  h)2  (y  k )2  r 2 , where (h, k) is centre of this circle and r be the radius. The centre of circle is the radical centre of the given circles and r is the length of tangent from (h, k) to any of the given three circles. The gradient of the radical axis of the circles x 2  y 2  3 x  4 y  5  0 and 3 x 2  3 y 2  7 x  8 y  11  0 is 1 3 U Example : 46 (a)  1 10 (c)  1 2 (d)  2 3 Equation of radical axis is S 1  S 2  0 ST Solution : (b) (b) [MP PET 2000] S1  x 2  y 2  3x  4y  5  0 , S 2  x 2  y 2  7 8 y 11 x  0 3 3 3  Radical axis is 2 x  20 y  4  0. Hence, gradient of radical axis =  Example : 47 1 10 The equations of three circles are x 2  y 2  12 x  16 y  64  0, 3 x 2  3 y 2  36 x  81  0 and x 2  y 2  16 x  81  0. The coordinates of the point from which the length of tangents drawn to each of the three circles is equal [Rajasthan PET 2002] (a) Solution : (d)  33  , 2   4  (b) (2, 2) (c) 33    2,  4   (d) None of these The required point is the radical centre of the three given circles Now, S 1  S 2  0   16 y  37  0 , S 2  S 3  0  4 x  54  0 and S 3  S 1  0   4 x  16 y  17  0 108 Circle and System of Circles Solving these equations, we get x  Example : 48 54 37 , y 4 16  x  27 37  27 37  ,. Hence the required point is  , y . 2 16  2 16  The equation of the circle, which passes through the point (2a, 0) and whose radical axis is x  a with respect to the circle 2 x 2  y 2  a 2 , will be Solution : (a) [Rajasthan PET 1999] x  y  2ax  0 2 (b) Equation of radical axis is x  x  y  2ax  0 2 2 a  2 (c) x  y  2ay  0 2 2 2x  a  0 Equation of required circle is x 2  y 2  a 2   (2 x  a)  0  It is passes through the point (2a, 0) ,  4 a 2  a 2   (4 a  a)  0  2 2 4.23 Co-Axial System of Circles. 2 2 E3 2 x  y  2ay  0 2   a  Equation of circle is x  y  a  2ax  a  0  x  y  2ax  0 2 (d) 2 60 (a) 2 A system (or a family) of circles, every pair of which have the same radical axis, are called co-axial circles. (1) The equation of a system of co-axial circles, when the equation of the radical axis and of one circle of the system are P  lx  my  n  0 and S+P=0 ID S+P=0 S+P=0 D YG U S=0 S  x 2  y 2  2 gx  2 fy  c  0 respectively, is S  P  0 ( is an arbitrary P=0 constant). (2) The equation of a co-axial system of circles, where the equation of any two circles of the system are S1+S2=0 S1=0 S1+( S1–S2)=0 S1+( S1+S2)=0 S1+( S1–S2)=0 S2=0 S2=0 S1=0 S1–S2=0 S 1  x 2  y 2  2 g1 x  2 f1 y  c1  0 and S 2  x 2  y 2  2 g 2 x  2 f2 y  c 2  0 Respectively, is S 1   (S 1  S 2 )  0 , (  1) or S 2  1 (S 1  S 2 )  0 , (1  1) Other form S 1  S 2  0, (  1) U (3) The equation of a system of co-axial circles in the simplest form is x 2  y 2  2 gx  c  0 , where g is variable and c is a constant. ST 4.24 Limiting Points. Limiting points of a system of co-axial circles are the centres of the point circles belonging to the family (Circles whose radii are zero are called point circles). (1) Limiting points of the co-axial system : Let the circle is x 2  y 2  2 gx  c  0 …..(i) where g is variable and c is constant.  Centre and the radius of (i) are ( g, 0) and (g 2  c) respectively. Let g2  c  0  g   c Thus we get the two limiting points of the given co-axial system as ( c , 0) and ( c , 0) Clearly the above limiting points are real and distinct, real and coincident or imaginary according as c>, =, 0. Then the circles are [Karnataka CET 1999] (a) Orthogonal Solution : (d) (b) Touching type (c) Intersecting type (d) Non intersecting type The equation of a system of circle with its centre on the axis of x is x 2  y 2  2 gx  c  0. Any point on the radical axis is (0, y 1 ) 60 Putting, x = 0, y    c If c is positive (c >0), we have no real point on radical axis, then circles are said to be non-intersecting type. 4.25 Image of the Circle by the Line Mirror. E3 Let the circle be x 2  y 2  2 gx  2 fy  c  0 and line mirror lx  my  n  0. In this condition, radius of circle remains unchanged but centre changes. Let the centre of image circle be ( x 1 , y 1 ). Slope of C1 C 2  slope of lx  my  n  1 …..(i) and mid point of C 1 ( g,  f ) and C 2 (x 1 , y 1 ) lie on lx  my  n  0 Solving (i) and (ii), we get ( x 1 , y 1 ) …..(ii) ID x g y  f  i.e., l  1  m  1 n  0  2   2  (–g,–f) r C1 r C2 Image circle Given circle lx+my+n=0 The equation of the image of the circle x 2  y 2  16 x  24 y  183  0 by the line mirror 4 x  7 y  13  0 is (a) (c) Solution : (d) x 2  y 2  32 x  4 y  235  0 D YG Example : 51 U  Required image circle is (x  x 1 )2  (y  y 1 )2  r 2 , where r  (g 2  f 2  c) x 2  y 2  32 x  4 y  235  0 (b) x 2  y 2  32 x  4 y  235  0 (d) x 2  y 2  32 x  4 y  235  0 The given circle and line are x 2  y 2  16 x  24 y  183  0 …..(i) and 4 x  7 y  13  0 …..(ii) Centre and radius of circle (i) are (– 8, 12) and 5 respectively. Let the centre of the image circle be (x 1 , y 1 ). Then slope of C1 C 2  slope of 4 x  7 y  13  1  y  12   4        1 or 4 y1  48  7 x 1  56   1   x1  8   7  or 7 x 1  4 y 1  104  0 (–8,12) …..(iii) U  x  8 y 1  12  ,  lie on 4 x  7 y  13  0 , and mid point of C1 C 2 i.e.,  1 2   2 ST  x  8   y1  12  then 4  1   7   13  0 or 4 x 1  7 y1  78  0  2   2  (x1,y1) 5 C2 C1 5 4x+7y+13=0 …..(iv) Solving (iii) and (iv), we get (x 1 , y 1 )  (16,  2)  Equation of the image circle is (x  16 ) 2  (y  2) 2  5 2 or x 2  y 2  32 x  4 y  235  0 4.26 Some Important Results. (1) Concyclic points : If A, B, C, D are concyclic then OA. OD  OC. OB , where O  be the centre of the circle. (2) Equation of the straight line joining two points  and  on the circle x y a 2 2 2             Required equation is x cos     y sin    a cos  2 2      2  A B O O D C Circle and System of Circles 111 (3) The point of intersection of the tangents at the point P () and Q () on the circle x 2  y 2  a 2 is            a cos   a sin    2 ,  2              cos   cos    2   2   Q() a a O …..(i) The centre and radius of the circle are P The maximum and minimum distance from P (x 1 , y 1 ) to the circle (i) are r  (g 2  f 2  c) 2 LR C(–g,–f) B ID PB  CB  PC  r  PC and PA  | CP  CA |  | PC  r | (P inside or outside) (5) Length of chord of contact is AB  (R 2  L2 ) R O L U D YG RL3 R 2  L2 S=0 A and area of the triangle formed by the pair of tangents and its chord of contact is  P(x1,y1) A E3 (g 2  f 2  c) respectively.  60 Let any point P (x 1 , y 1 ) and circle x 2  y 2  2 gx  2 fy  c  0 where P( )  (4) Maximum and Minimum distance of a point from the circle : C ( g,  f ) and – M R L B Where R is the radius of the circle and L is the length of tangent from P (x 1 , y 1 ) on S=0. Here L  S 1. (6) Length of an external common tangent and internal common tangent to two circles is given by Length of external common tangent Lex  d  (r1  r2 ) 2 A 2 Lin  d  (r1  r2 ) U 2 d r1 and length of internal common tangent 2 Lex (r1–r2) r1+r2 [Applicable only when d  (r1  r2 ) ] r1 Lin r2 A B B r2 r 2 C2 Lin A ST where d is the distane between the centres of two circles i.e., | C 1 C 2 |  d and r1 and r2 are the radii of two circles. (7) Family of circles circumscribing a triangle whose sides are given by L1  0; L 2  0 and L 3  0 is given by L1 L 2  L 2 L 3  L 3 L1  0 provided coefficient of xy  0 and coefficient of x 2  coefficient of y 2. Equation of the circle circumscribing the triangle formed by the lines a r x  b r y  c r  0, where r = 1, 2, 3, is a12  b12 a1 x  b1 y  c1 a 22  b 22 a2 x  b 2 y  c 2 a 32  b 32 a3 x  b 3 y  c 3 a1 b1 a2 b2  0 A L1=0 B a3 b3 L3=0 L2=0 C 112 Circle and System of Circles (8) Equation of circle circumscribing a quadrilateral whose sides in order are represented by the lines L1  0, L 2  0, L 3  0 and L 4  0 is given by A L1 L3  L 2 L4  0 L3=0 D L4=0 provided coefficient of x 2  coefficient of y 2 and coefficient of xy  0 B L2=0 L1=0 60 C (9) Equation of the circle circumscribing the triangle PAB is A (x  x 1 ) (x  g)  (y  y 1 ) (y  f )  0 E3 O(0,0) where O ( g,  f ) is the centre of the circle x 2  y 2  2 gx  2 fy  c  0 P(–8,2) B (10) Locus of mid point of a chord of a circle x 2  y 2  a 2 , which subtends an angle  at the centre is ID x 2  y 2  (a cos  / 2) 2 (11) The locus of mid point of chords of circle x 2  y 2  a 2 , which are making right angle at centre is a2 2 U x2  y2  (12) The locus of mid point of chords of circle x 2  y 2  2 gx  2 fy  c  0 , which are making right angle at D YG origin is x 2  y 2  gx  fy  c / 2  0. (13) The area of triangle , which is formed by co-ordinate axes and the tangent at a point ( x 1 , y 1 ) of circle x 2  y 2  a 2 is a 4 / 2 x 1 y 1 (14) If a point is outside, on or inside the circle then number of tangents from the points is 2, 1 or none. U (15) A variable point moves in such a way that sum of square of distances from the vertices of a triangle remains constant then its locus is a circle whose centre is the centroid of the triangle. (16) If the points where the line a1 x  b1 y  c1  0 and a 2 x  b 2 y  c 2  0 meets the coordinate axes are ST concyclic then a1 a 2  b1 b 2. Example : 52  1   , i  1, 2, 3, 4 are concylic points, then the value of m1. m2. m3. m4 is If  m i , m i   (b) – 1 (a) 1 Solution : (a) (c) 0 [Karnataka CET 2002] (d) None of these Let the equation of circle be x  y  2 gx  2 fy  c  0 2 2  1   lies on this circle Since the point  m i ,  m i   2  mi  1 mi 2  2 gm i  2f  c  0  m i 4  2 gm i 3  cm i 2  2 fm i  1  0 mi Clearly its roots are m 1 , m 2 , m 3 and m 4 ,  m1. m2. m3. m4 = product of roots  Example : 53 1 1 1 Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals [IIT Screening 2001] Circle and System of Circles 113 (a) Solution : (a) (b) PQ. RS tan   PQ  RS 2 (c) 2 PQ. RS PQ  RS PQ PQ  PR 2r /2  /2- 60 R PQ. RS 4r2 r C ID E3  4 r 2  PQ. RS  2r  (PQ) (RS ) D YG U *** U Q X RS i.e. cot   2r ST PQ 2  RS 2 2 S   RS Also tan      2   2r  tan . cot   (d) r P