Circles PDF

Nature of Chapter: 1. You should know all the formulae, you covered in straight lines, before starting this chapter. 2. You will learn some standard approaches and notations in this chapter without which you cannot proceed to Conics. 3. There are very few formulae in this chapter. Its majorl...

Nature of Chapter: 1. You should know all the formulae, you covered in straight lines, before starting this chapter. 2. You will learn some standard approaches and notations in this chapter without which you cannot proceed to Conics. 3. There are very few formulae in this chapter. Its majorly about basic geometry of circle and some solving techniques.  Weightage of Circles (Last 5 years) 2023 2022 2021 2020 2019 Average JEE Main 2.5 % 3.5 % 3.7 % 1.8 % 4.0 % 3.10 % Jee Advanced 12 % 6% 13 % 3% 7% 8.20 %  Circles Standard Equations of a Circle Intercepts made by a Circle Some Standard Notations Position of a Point with respect to a Circle Position of a Line with respect to a Circle and Equations of Tangents Common Tangents of Circles Family of Circles Chords of a Circle Orthogonality of Two Circles Radical Axis and Radical Centre  Circles Critical Topics in the Chapter Standard equations of circles Standard Notations Equation of tangents Common Tangents Fill The Google Form available in Description https://docs.google.com/forms/d/1atjRl0T6zDkK- 6YnL5J5Us_5MaM8U8DZYn2B4TIWdrA/edit Standard equations of a circle  Standard equations of a circle Central form of the Equation of a Circle Here, centre is (x1, y1) and radius is ‘r’.  Standard equations of a circle Central form of the Equation of a Circle Here, centre is (x1, y1) and radius is ‘r’. NOTE Circle with centre at (0, 0) and radius r is x 2 + y 2 = r 2  Standard equations of a circle Central form of the Equation of a Circle For example, (4, 7) (a) (1, 3) (b) 4 (2, 2)  Standard equations of a circle Central form of the Equation of a Circle For example, (4, 7) centre is (1, 3) and =5 (a) So, the circle’s equation is (1, 3) (x − 1)2 + (y − 3)2 = 25, that is x2 + y2 − 2x − 6y − 15 = 0 centre is (2, 2) and circumference = 4𝜋 (b) ⇒ 2𝜋r = 4𝜋 ⇒ r = 2 (2, 2) 4 So, circle’s equation is (x − 2)2 + (y − 2)2 = 4, that is x2 + y2 − 4x − 4y + 4 = 0  Standard equations of a circle Remark The diameter or the normal of a circle passes through its centre. Q The lines 4x −y = 7 and 2x − 5y = 8 are diameters of a circle of area 154 sq. units. The equation of this circle is _____. (take π = 22/7) Q The lines 4x −y = 7 and 2x − 5y = 8 are diameters of a circle of area 154 sq. units. The equation of this circle is _____. (take π = 22/7) Solution: Q A circle has radius 3 units and its centre lies on the line y = x - 1. Find the equation of the circle if it passes through (7, 3).  Q A circle has radius 3 units and its centre lies on the line y = x - 1. Find the equation of the circle if it passes through (7, 3). Solution: C (t, t - 1) P(7, 3) y=x-1 Q Find the equation of circle which passes through the points (1, -2), (4, -3) and whose centre lies on the line x + y = 2. Remark Perpendicular of any chord of the circle, passes through center of circle. Q Find the equation of circle which passes through the points (1, -2), (4, -3) and whose centre lies on the line x + y = 2. Solution: Q Find the equation of circle which passes through the points (1, -2), (4, -3) and whose centre lies on the line x + y = 2. Solution:  Standard equations of a circle General form of the Equation of a circle Recall A general two degree equation in two variables is represented as  Standard equations of a circle General form of the Equation of a circle x2 + y2 + 2gx + 2fy + c = 0 NOTE Condition for a general second degree equation in two variables to represent a circle is a = b and h = 0. If a = b ≠ 1, then we divide the equation by a constant to make both coefficients equal to 1.  Standard equations of a circle General form of the Equation of a circle x2 + y2 + 2gx + 2fy + c = 0 where, Centre is (-g , -f) and For example: Centre Radius x2 + y2 - 4x + 4y - 28 = 0  Standard equations of a circle General form of the Equation of a circle x2 + y2 + 2gx + 2fy + c = 0 where, Centre is (-g , -f) and For example: Centre Radius x2 + y2 - 4x + 4y - 28 = 0 (2, -2) 6 (-3/4, -1) 1  Equation of the circle concentric with the circle Q x2 + y2 - 6x + 12y + 15 = 0 and of double its area is A x2 + y2 - 3x + 12y -15 = 0 B x2 + y2 - 3x + 12y -30 = 0 C x2 + y2 - 6x + 12y - 15 = 0 D x2 + y2 - 6x + 12y -20 = 0  Equation of the circle concentric with the circle Q x2 + y2 - 6x + 12y + 15 = 0 and of double its area is A x2 + y2 - 3x + 12y -15 = 0 B x2 + y2 - 3x + 12y -30 = 0 C x2 + y2 - 6x + 12y - 15 = 0 D x2 + y2 - 6x + 12y -20 = 0  Equation of the circle concentric with the circle Q x2 + y2 - 6x + 12y + 15 = 0 and of double its area is Solution: and centre is (3, -6) Q Find the equation of the circle which passes through the points (5, -8), (2, -9) and (2, 1). Q Find the equation of the circle which passes through the points (5, -8), (2, -9) and (2, 1). Solution: Let the equation of required circle be Since, it passes through (5, -8), (2, -9) and (2, 1). On substituting the coordinates of three points in equation (i), we get On solving (iii), (iv) and (v), we get f = 4, g = -2 and c = - 5 Therefore, equation of the required circle is x2 + y2 - 4x + 8y - 5 = 0  Single correct question JEE Advanced 2021, P1 Q Consider a triangle Δ whose two sides lie on the x-axis and the line x + y + 1 = 0. If the orthocenter of Δ is (1, 1), then the equation of the circle passing through the vertices of the triangle Δ is A x2 + y2 - 3x + y = 0 B x2 + y2 + x + 3y = 0 C x2 + y2 + 2y - 1 = 0 D x2 + y2 + x + y = 0  Single correct question JEE Advanced 2021, P1 Q Consider a triangle Δ whose two sides lie on the x-axis and the line x + y + 1 = 0. If the orthocenter of Δ is (1, 1), then the equation of the circle passing through the vertices of the triangle Δ is A x2 + y2 - 3x + y = 0 B x2 + y2 + x + 3y = 0 C x2 + y2 + 2y - 1 = 0 D x2 + y2 + x + y = 0 Solution: x+y+1=0 B(1, -2) H(1, 1) y=0 A(-1, 0) C(0, 0)  Standard equations of a circle Some Special Circles Now let us see some special cases. These are generally used to give information indirectly, in the questions.  Standard equations of a circle Some Special Circles 1. Circle touching X - axis 2. Circle touching Y - axis Y (0, b) X (a, 0)  Standard equations of a circle Some Special Circles 3. Circle touching X - axis at origin 4. Circle touching Y - axis at origin Y Y X O X O  Standard equations of a circle Some Special Circles 5. Circle touching both the axes Y Y O X O X Y Y X O X O  JEE Main 2013 Q The circle passing through (1, −2) and touching the X - axis at the point (3, 0) also passes through the point _____. A (5, 2) B (-5, 2) C (5, -2) D (-5, -2)  JEE Main 2013 Q The circle passing through (1, −2) and touching the X - axis at the point (3, 0) also passes through the point _____. A (5, 2) B (-5, 2) C (5, -2) D (-5, -2)  JEE Main 2013 Q The circle passing through (1, −2) and touching the X - axis at the point (3, 0) also passes through the point _____. Solution:  ABCD is a square of unit area. A circle is touching two Q sides of ABCD and passes through one of its vertices. Find radius of the circle.  ABCD is a square of unit area. A circle is touching two Q sides of ABCD and passes through one of its vertices. Find radius of the circle. Solution: Y C (1, 1) D (0, 1) a (a, a) X A (0, 0) B (1, 0)  Consider circles C1 & C2 touching both the axes Q and passing through (4, 4), then the product of radii of these circles is  Consider circles C1 & C2 touching both the axes Q and passing through (4, 4), then the product of radii of these circles is Ans : 32  Consider circles C1 & C2 touching both the axes Q and passing through (4, 4), then the product of radii of these circles is Solution: Since circle touches both the axes and passes through (4, 4), it lies in the first quadrant, so its equation is  Standard equations of a circle Diametric form of the Equation of a Circle The circle whose diameter endpoints are A(x1, y1) and B(x2, y2) has equation (x − x1) (x − x2) + (y − y1) (y − y2) = 0  Standard equations of a circle Diametric form of the Equation of a Circle The circle whose diameter endpoints are A(x1, y1) and B(x2, y2) has equation (x − x1) (x − x2) + (y − y1) (y − y2) = 0 NOTE Basically it’s the sum of two quadratics, one in x, whose roots are the abscissae and one in y, whose roots are the ordinates of the diametric endpoints.  AIEEE 2011 Q Find the equation of the circle which passes through (1, 0) and (0, 1) and has its radius as small as possible.  AIEEE 2011 Q Find the equation of the circle which passes through (1, 0) and (0, 1) and has its radius as small as possible. Solution: The radius will be minimum, if the given points are the end-points of a diameter. Then, equation of circle is (x - 1) (x - 0) + (y - 0) (y - 1) = 0 ⇒ x2 + y2 - x - y = 0  Standard equations of a circle Parametric form of the Equation of a Circle Y r θ X O  Standard equations of a circle Parametric form of the Equation of a Circle Y (a) x2 + y2 = r2 ⇒ x = r cos θ, y = r sin θ P(θ) r θ X O In particular, a general point on x2 + y2 = 1 is of the form (cosθ, sinθ) for some θ.  Standard equations of a circle Parametric form of the Equation of a Circle (b) (x − x1)2 + (y − y1)2 = r2 ⇒ x = x1 + r cosθ, y = y1 + r sin θ Q The parametric equations of the circle x2 + y2 + mx + my = 0 are A B C x = 0, y = 0 D None of these Q The parametric equations of the circle x2 + y2 + mx + my = 0 are A B C x = 0, y = 0 D None of these Q The parametric equations of the circle x2 + y2 + mx + my = 0 are Solution: So, radius = and centre = , Q If A(cos ⍺, sin ⍺), B(cos β, sin β), C(cos γ, sin γ) are the vertices of a ΔABC, then find the coordinates of its orthocentre. Q If A(cos ⍺, sin ⍺), B(cos β, sin β), C(cos γ, sin γ) are the vertices of a ΔABC, then find the coordinates of its orthocentre. Solution: Q If A(cos ⍺, sin ⍺), B(cos β, sin β), C(cos γ, sin γ) are the vertices of a ΔABC, then find the coordinates of its orthocentre. Solution: Intercepts made by a circle  Intercepts made by a circle y = mx + c A AB is the intercept made by circle on the line y = mx + c.  Intercepts made by a circle y = mx + c A AB is the intercept made by circle on the line y = mx + c. NOTE Whenever a circle makes an intercept on line, always refer to following figure. r Q Find the equation of circle having centre at (3, − 1) and cutting an intercept of length 6 units on the line 2x − 5y + 18 = 0. A (x − 3)2 + (y + 1)2 = 31 B (x − 3)2 + (y + 1)2 = 29 C (x − 3)2 + (y + 1)2 = 38 D None of these Q Find the equation of circle having centre at (3, − 1) and cutting an intercept of length 6 units on the line 2x − 5y + 18 = 0. A (x − 3)2 + (y + 1)2 = 31 B (x − 3)2 + (y + 1)2 = 29 C (x − 3)2 + (y + 1)2 = 38 D None of these  Q Find the equation of circle having centre at (3, − 1) and cutting an intercept of length 6 units on the line 2x − 5y + 18 = 0. Solution: P M B  Multiple choice question JEE Advanced 2013 Circle(s) touching x-axis at a distance 3 from the Q origin and having an intercept of length on y-axis (are) A x2 + y2 - 6x + 8y + 9 = 0 B x2 + y2 - 6x + 7y + 9 = 0 C x2 + y2 - 6x - 8y + 9 = 0 D x2 + y2 - 6x - 7y + 9 = 0  Multiple choice question JEE Advanced 2013 Circle(s) touching x-axis at a distance 3 from the Q origin and having an intercept of length on y-axis (are) A x2 + y2 - 6x + 8y + 9 = 0 B x2 + y2 - 6x + 7y + 9 = 0 C x2 + y2 - 6x - 8y + 9 = 0 D x2 + y2 - 6x - 7y + 9 = 0  Multiple choice question JEE Advanced 2013 Circle(s) touching x-axis at a distance 3 from the Q origin and having an intercept of length on y-axis (are) Solution: (3, 𝞪) (3, 0)  Intercepts made by a circle Remark 1. Intercept made by x2 + y2 + 2gx + 2fy + c = 0 on the X - axis A B X  Intercepts made by a circle Remark 1. Intercept made by x2 + y2 + 2gx + 2fy + c = 0 on the X - axis A B X (a) g2 − c > 0 ⇒ Circle cuts the X - axis at two distinct points (b) g2 − c = 0 ⇒ Circle touches the X - axis (c) g2 − c < 0 ⇒ Circle does not meet the X - axis  Intercepts made by a circle Remark 2. Intercept made by x2 + y2 + 2gx + 2fy + c = 0 on the Y - axis Y B A  Intercepts made by a circle Remark 2. Intercept made by x2 + y2 + 2gx + 2fy + c = 0 on the Y - axis Y B A (a) f 2 − c > 0 ⇒ Circle cuts the Y - axis at two distinct points (b) f 2 − c = 0 ⇒ Circle touches the Y - axis (c) f 2 − c < 0 ⇒ Circle does not meet the Y - axis  Find equation of locus of centre of circle which Q touches Y - axis and having intercept on X-axis of length 2l. A x2 + y2 = l 2 B x2 - y2 = l 2 C x2 + y2 = 4l 2 D x2 - y2 = 4l 2  Find equation of locus of centre of circle which Q touches Y - axis and having intercept on X-axis of length 2l. A x2 + y2 = l 2 B x2 - y2 = l 2 C x2 + y2 = 4l 2 D x2 - y2 = 4l 2  Find equation of locus of centre of circle which Q touches Y - axis and having intercept on X-axis of length 2l. Solution: Y B (h, k) X O D l A 2l Q 2 rods whose lengths are 2a, 2b slide along axes (one on each) in such a way that their extremities are always concyclic. Find the equation of locus of centre of circle. Solution: Y D r 2b F O (h, k) r C X A B E 2a Alternate Solution Y D r 2b F O (h, k) r C X A B E 2a Some standard Notations  Some standard Notations Here, we will be learning some standards notations for general second degree equations in x and y. These notations will be very helpful in upcoming formulae. Primarily there are are three notations S, S1 and T. Let’s see what do they denote.  Some standard Notations Notations: Any second degree equation in two variables, that is, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 will be represented as S = 0. As of now, that we are doing circles, so we have S ≡ x2 + y2 + 2gx + 2fy + c  Some standard Notations 1. S ≡ x2 + y2 + 2gx + 2fy + c 2. Consider a point (x1, y1). Value of S at (x1, y1) is represented by S1 i.e., S1 = x12 + y12 + 2gx1 + 2fy1 + c 3. Consider a point (x1, y1). If in S we replace then we get T,  Write S1 and T for the following: Q (a) S ≡ x2 + 2y2 − 3x + 4y + 3 at the point (1, 2) (b) S ≡ x2 + 2xy +3y at the point (1, 3)  Write S1 and T for the following: Q (a) S ≡ x2 + 2y2 − 3x + 4y + 3 at the point (1, 2)  Write S1 and T for the following: Q (a) S ≡ x2 + 2y2 − 3x + 4y + 3 at the point (1, 2) Solution:  Write S1 and T for the following: Q (b) S ≡ x2 + 2xy +3y at the point (1, 3)  Write S1 and T for the following: Q (b) S ≡ x2 + 2xy +3y at the point (1, 3) Solution: Position of a point w.r.t. a Circle  Position of a point w.r.t. a circle Method 1 Find distance of point P from centre of circle O. OP < r ⇒ P lies inside the circle OP = r ⇒ P lies on the circle OP > r ⇒ P lies outside the circle  Position of a point w.r.t. a circle Method 1 Find distance of point P from centre of circle O. OP < r ⇒ P lies inside the circle OP = r ⇒ P lies on the circle OP > r ⇒ P lies outside the circle Method 2 S1 < 0 ⇒ P lies inside the circle S1 = 0 ⇒ P lies on the circle S1 > 0 ⇒ P lies outside the circle  Position of a point w.r.t. a circle Remark Greatest and least distance of a point from a circle. P |OP - r| = least distance of point P from the circle |OP + r| = greatest distance of point P from the circle Q Find the shortest and the longest distance from the point (2, -7) to the circle x2 + y2 - 14x - 10y - 151 = 0. Q Find the shortest and the longest distance from the point (2, -7) to the circle x2 + y2 - 14x - 10y - 151 = 0. Solution: Position of a Line w.r.t. a Circle and Equation of Tangents  Position of a Line w.r.t. a circle and Equation of Tangents For a given line and a circle, either (a)line cuts the circle, or (b)line touches the circle, or (c) line does not meet the circle In the section we will be studying their conditions. No doubt, major focus will be (b), that is tangency condition.  Position of a Line w.r.t. a circle and Equation of Tangents Method 1 Find distance d of centre of circle from given line d < r ⇒ line cuts the circle d = r ⇒ line is tangent to circle d > r ⇒ line does not meet circle  Position of a Line w.r.t. a circle and Equation of Tangents Method 2 Solve line with circle to get a quadratic equation. D > 0 ⇒ line cuts the circle D = 0 ⇒ line is tangent to circle D < 0 ⇒ line does not meet circle  Position of a Line w.r.t. a circle and Equation of Tangents Result 1. Equations of tangents to x2 + y2 = r2, having slope m are E.g. Tangents to x2 + y2 = 4, having slope 3 are ___________.  Position of a Line w.r.t. a circle and Equation of Tangents Result 2. Equations of tangents to (x − x1)2 + (y − y1)2 = r2, having slope m are E.g. Tangents to the circle (x − 1)2 + (y − 5)2 = 4, having slope 3 are__________.  (a) Find 𝜆 such that y = 2x + 𝜆 is tangent to x2 + y2 = 5. Q (b) Find 𝜆 so that 3x − 4y = 𝜆 is tangent to x2 + y2 − 4x − 8y − 5 = 0. Q (a) Find 𝜆 such that y = 2x + 𝜆 is tangent to x2 + y2 = 5. Q (a) Find 𝜆 such that y = 2x + 𝜆 is tangent to x2 + y2 = 5. Solution: Alternate Solution Q (b) Find 𝜆 so that 3x − 4y = 𝜆 is tangent to x2 + y2 − 4x − 8y − 5 = 0. Q (b) Find 𝜆 so that 3x − 4y = 𝜆 is tangent to x2 + y2 − 4x − 8y − 5 = 0. Solution: Q If the line lx + my + n = 0 is tangent to the circle x2 + y2 = a2, then find the condition. Q If the line lx + my + n = 0 is tangent to the circle x2 + y2 = a2, then find the condition. Solution: Q The value of A for which the set {(x, y) : x2 + y2 - 6x + 4y ≤ 12} ∩ {(x, y) : 4x + 3y ≤ λ} contains only one point is A 31 B -31 C 19 D -19 Q The value of A for which the set {(x, y) : x2 + y2 - 6x + 4y ≤ 12} ∩ {(x, y) : 4x + 3y ≤ λ} contains only one point is A 31 B -31 C 19 D -19 Q The value of A for which the set {(x, y) : x2 + y2 - 6x + 4y ≤ 12} ∩ {(x, y) : 4x + 3y ≤ λ} contains only one point is Solution: Q Find equations of tangents to x2 + y2 = 16 drawn from (1, 4). Q Find equations of tangents to x2 + y2 = 16 drawn from (1, 4). Solution: (1, 4)  The locus of the point of intersection of the Q perpendicular tangents to the circle x2 + y2 = a2 is ___. A x2 + y2 = 2a2 B x2 + y2 = 3a2 C x2 + y2 = 5a2 D None of these  The locus of the point of intersection of the Q perpendicular tangents to the circle x2 + y2 = a2 is ___. A x2 + y2 = 2a2 B x2 + y2 = 3a2 C x2 + y2 = 5a2 D None of these  The locus of the point of intersection of the Q perpendicular tangents to the circle x2 + y2 = a2 is ___. Solution: A O P (h, k)  Position of a Line w.r.t. a circle and Equation of Tangents NOTE Locus of point of intersection of perpendicular tangents to a circle is called the Director circle. In case of circle, the director circle is the concentric circle with radius equal to √2 times the radius of the given circle.  Position of a Line w.r.t. a circle and Equation of Tangents NOTE Locus of point of intersection of perpendicular tangents to a circle is called the Director circle. In case of circle, the director circle is the concentric circle with radius equal to √2 times the radius of the given circle. E.g. Director circle of (x − 1)2 + (y − 2)2 = 3 is ___________.  Position of a Line w.r.t. a circle and Equation of Tangents NOTE Locus of point of intersection of perpendicular tangents to a circle is called the Director circle. In case of circle, the director circle is the concentric circle with radius equal to √2 times the radius of the given circle. E.g. Director circle of (x − 1)2 + (y − 2)2 = 3 is (x − 1)2 + (y − 2)2 = 6.  The tangents to x2 + y2 = a2 having inclinations Q α and β intersect at P. If cot α + cot β = 0, then the locus of P is A x+y=0 B x-y=0 C xy = 0 D None of these  The tangents to x2 + y2 = a2 having inclinations Q α and β intersect at P. If cot α + cot β = 0, then the locus of P is A x+y=0 B x-y=0 C xy = 0 D None of these  The tangents to x2 + y2 = a2 having inclinations Q α and β intersect at P. If cot α + cot β = 0, then the locus of P is Solution: x2 + y2 = a2  Position of a Line w.r.t. a circle and Equation of Tangents So far, we have studied (1) condition for tangency (2) slope form of tangent We do have other forms of equations of tangents as well.  Various Equation of Tangents Slope form Tangent at a Point on a Circle Parametric form slope = m (x1, y1) P(θ) x2 + y2 = x2 + y2 = r2 r2  Various Equation of Tangents Slope form Tangent at a Point on a Circle Parametric form slope = m (x1, y1) P(θ) x2 + y2 = x2 + y2 = r2 r2 T=0 x cosθ + y sinθ = r Q Tangent to circle x2 + y2 = 5 at (1, -2) also touches the circle x2 + y2 - 8x + 6y + 20 = 0. Find the coordinates of the corresponding point of contact Q Tangent to circle x2 + y2 = 5 at (1, -2) also touches the circle x2 + y2 - 8x + 6y + 20 = 0. Find the coordinates of the corresponding point of contact Solution: Equation of tangent to x2 + y2 = 5 at (1, -2) is x - 2y - 5 = 0 Putting x = 2y + 5 in second circle, we get (2y + 5)2 + y2 - 8(2y + 5) + 6y + 20 = 0 ⇒ 5y2 + 10y + 5 = 0 ⇒ y = -1 ⇒ x = -2 + 5 = 3 Thus, point of contact is (3, -1).  Multiple correct question JEE Advanced 2016, P1 Let RS be the diameter of the circle x2 + y2 = 1, where S ≡ (1, 0). Let Q P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point (s) A B C D  Multiple correct question JEE Advanced 2016, P1 Let RS be the diameter of the circle x2 + y2 = 1, where S ≡ (1, 0). Let Q P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point (s) A B C D Solution: P(𝜃) Q E(h, k) S(1, 0)  Remark For a given circle S = 0, (1) Length of tangent from point P T P (2) S1 is also called power of point P(x1, y1) with respect to S = 0.  Observations If a pair of tangents is drawn from a point P to a circle as shown in the figure, then A R L P O R L B (1) angle between the pair of tangents is given by (2) Length of AB = (3) Area triangle PAB = (4) circumcircle of Δ PAB has OP as diameter.  A R L O P R L B  JEE Main 2020 Let the tangents drawn from the origin to the circle, Q x2 + y2 – 8x – 4y + 16 = 0 touch it at the point A and B. Then (AB)2 is equal to: A B C D  JEE Main 2020 Let the tangents drawn from the origin to the circle, Q x2 + y2 – 8x – 4y + 16 = 0 touch it at the point A and B. Then (AB)2 is equal to: A B C D  JEE Main 2020 Let the tangents drawn from the origin to the circle, Q x2 + y2 – 8x – 4y + 16 = 0 touch it at the point A and B. Then (AB)2 is equal to: Solution: Common Tangents of circles  Common Tangents of circles In this topic, we will try to observe the number of common tangents of two circles, depending upon their positions. We will also learn how to find lengths and equations of the common tangents.  Common Tangents of circles Try to observe how we can comment upon the positions of two circles depending on their radii and the distance between their centres. r1 r2 r1 r2 (1) C1 (2) C1 (3) C1 C2 C2 C2 (4) C2 (5) C2 C1 C1  Common Tangents of circles Try to observe how we can comment upon the positions of two circles depending on their radii and the distance between their centres. r1 r2 r1 r2 (1) C1 (2) C1 (3) C1 C2 C2 C2 (4) C2 (5) C2 C1 C1 Number of Common Tangents TCT DCT (1) |C1C2| > r1 + r2 ⇒ (2) |C1C2| = r1 + r2 ⇒ (3) |r1 - r2| < |C1C2| < r1 + r2 ⇒ (4) |C1C2| = |r1 - r2| ⇒ (5) |C1C2| < |r1 + r2| ⇒ Number of Common Tangents TCT DCT (1) |C1C2| > r1 + r2 ⇒ 4 common tangents (2) |C1C2| = r1 + r2 3 common tangents ⇒ (3) |r1 - r2| < |C1C2| < r1 + r2 ⇒ 2 common tangents (4) |C1C2| = |r1 - r2| ⇒ 1 common tangent (5) |C1C2| < |r1 + r2| ⇒ 0 common tangents  JEE Main 15th April 2023 Q The number of common tangents, to the circles x2 + y2 - 18x - 15y + 131 = 0 and x2 + y2 - 6x - 6y - 7 = 0, is A 4 B 2 C 3 D 1  JEE Main 15th April 2023 Q The number of common tangents, to the circles x2 + y2 - 18x - 15y + 131 = 0 and x2 + y2 - 6x - 6y - 7 = 0, is A 4 B 2 C 3 D 1  JEE Main 15th April 2023 Q The number of common tangents, to the circles x2 + y2 - 18x - 15y + 131 = 0 and x2 + y2 - 6x - 6y - 7 = 0, is Solution:  If the circles x2 + y2 - 10x + 16y + 89 - r2 = 0 and Q x2 + y2 + 6x - 14y + 42 = 0 have common points, then the number of possible integral values of r is equal to A 13 B 14 C 15 D 18  If the circles x2 + y2 - 10x + 16y + 89 - r2 = 0 and Q x2 + y2 + 6x - 14y + 42 = 0 have common points, then the number of possible integral values of r is equal to A 13 B 14 C 15 D 18  If the circles x2 + y2 - 10x + 16y + 89 - r2 = 0 and Q x2 + y2 + 6x - 14y + 42 = 0 have common points, then the number of possible integral values of r is equal to Solution:  JEE Main 2014 Q Let C be the circle centred at (1, 1) and having radius 1. If T is the circle centred at (0, y) passing through the origin and touching the circle C externally, then the radius of T is equal to ____. A B C D  JEE Main 2014 Q Let C be the circle centred at (1, 1) and having radius 1. If T is the circle centred at (0, y) passing through the origin and touching the circle C externally, then the radius of T is equal to ____. A B C D Solution:  Common Tangents of circles Remark r2 r1 C1 P (external) C2 r1 C2 C1 (internal) P r2  If area of triangle formed by common tangents Q to the circles x2 + y2 - 6x = 0 and x2 + y2 + 2x = 0 is then value of λ is_____.  If area of triangle formed by common tangents Q to the circles x2 + y2 - 6x = 0 and x2 + y2 + 2x = 0 is then value of λ is_____. Ans: 3.00  If area of triangle formed by common tangents Q to the circles x2 + y2 - 6x = 0 and x2 + y2 + 2x = 0 is then value of λ is_____. Solution:  Common Tangents of circles Lengths of Common Tangents (1) Direct Common Tangent T1 T2 C1 C2 (2) Transverse Common Tangent T1 C2 C1 T2  Common Tangents of circles Lengths of Common Tangents (1) Direct Common Tangent T1 T2 r1 − r2 l C1 C2 (2) Transverse Common Tangent T1 C2 C1 l T2 r1 + r2 Family of Circles  Family of circles Just like family of lines, we have family of circles too, such as circles passing through intersection of two circles, a circle and a line etc.  Family of circles (1) S + L = 0 S=0 L=0 E.g. Any circle passing through x2 + y2 − 2x −3 = 0 and x − 2y + 1 = 0 is of the form ____________.  Find equation of circle passing through points of Q intersection of the line x + y − 1 = 0 and the circle x2 + y2 = 9 and which also passes through the point (3, 4).  Find equation of circle passing through points of Q intersection of the line x + y − 1 = 0 and the circle x2 + y2 = 9 and which also passes through the point (3, 4). Solution:  Family of circles (2) S + λS’ = 0, λ ≠ -1 S=0 S’ = 0  Family of circles (2) S + λS’ = 0, λ ≠ -1 S=0 S’ = 0 NOTE S - S’ = 0 gives the equation of the common chord of S = 0 and S’ = 0.  JEE Main 2020 Q The circle passing through the intersection of the circles, x2 + y2 - 6x = 0 and x2 + y2 - 4y = 0, having its centre on the line, 2x - 3y + 12 = 0, also passes through the point A (-3, 6) B (-1, 3) C (-3, 1) D (1, -3)  JEE Main 2020 Q The circle passing through the intersection of the circles, x2 + y2 - 6x = 0 and x2 + y2 - 4y = 0, having its centre on the line, 2x - 3y + 12 = 0, also passes through the point A (-3, 6) B (-1, 3) C (-3, 1) D (1, -3) Solution: Q If the circle x2 + y2 + 4x + 22y + c = 0 bisects the circumference of the circle x2 + y2 − 2x + 18y − d = 0, then find c + d. Q If the circle x2 + y2 + 4x + 22y + c = 0 bisects the circumference of the circle x2 + y2 − 2x + 18y − d = 0, then find c + d. Solution:  Family of circles (3) Family of circles passing through two given points A(x1, y1) and B(x2, y2) A (x1, y1) B (x2, y2) E.g. Any circle through (1, 1) and (2, 2) is of the form  Family of circles (3) Family of circles passing through two given points A(x1, y1) and B(x2, y2) Consider Then, S + 𝜆L = 0 gives family of circles passing through A(x1, y1) and B(x2, y2) E.g. Any circle through (1, 1) and (2, 2) is of the form (x − 1) (x − 2) + (y − 1) (y − 2) + 𝜆(y − x) = 0.  Family of circles (4) Family of circles tangent to a given line at a given point L=0 A (x1, y1)  Family of circles (4) Family of circles tangent to a given line at a given point L=0 A (x1, y1) Say that the given line and the point are respectively L = 0 and A (x1, y1). Consider, a circle through A (x1, y1), S : (x − x1)2 + (y − y1)2 Then S + 𝜆L = 0 gives a family of circles touching the line L = 0 at point A (x1, y1). E.g. Any circle touching x + y + 1 = 0 at (1, − 2) is of the form ___  Family of circles (4) Family of circles tangent to a given line at a given point L=0 A (x1, y1) Say that the given line and the point are respectively L = 0 and A (x1, y1). Consider, a circle through A (x1, y1), S : (x − x1)2 + (y − y1)2 Then S + 𝜆L = 0 gives a family of circles touching the line L = 0 at point A (x1, y1). E.g. Any circle touching x + y + 1 = 0 at (1, − 2) is of the form (x − 1)2 + (y + 2)2 + 𝜆(x + y + 1) = 0. Q If the radius of the circle passing through the origin and touching the line x + y = 2 at (1, 1) is r units, then the value of is Q If the radius of the circle passing through the origin and touching the line x + y = 2 at (1, 1) is r units, then the value of is Ans: 3 Q If the radius of the circle passing through the origin and touching the line x + y = 2 at (1, 1) is r units, then the value of is Solution:  JEE Main 2019 If a circle C passing through the point (4, 0) touches the Q circle x2 + y2 + 4x - 6y = 12 externally at the point (1, -1), then the radius of C is A 5 B C D 4  JEE Main 2019 If a circle C passing through the point (4, 0) touches the Q circle x2 + y2 + 4x - 6y = 12 externally at the point (1, -1), then the radius of C is A 5 B C D 4 Solution: B(4, 0) A (-2, 3) Chords of Circles  Chords of circles Here, we will be studying (1) Chord of contact (2) Chord with given midpoint  Chords of circles (1) Equation of CoC (chord of contact) with respect to P(x1, y1) Its equation is given by T = 0 P (x1, y1) S=0 (2) Equation of chord with given midpoint P(x1, y1) Its equation given by T = S1 P (x1, y1) S=0 Q If the straight line x - 2y + 1 = 0 intersects the circle x2 + y2 = 25 in points P and Q, then find the coordinates of the point of intersection of tangents drawn at P and Q to the circle x2 + y2 = 25. Solution: x - 2y + 1 = 0 P R Q  Tangents are drawn to the circle x2 + y2 = 16 at the points Q where it intersects the circle x2 + y2 - 6x - 8y - 8 = 0, then the point of intersection of these tangents is A (4, 16/3) B (12, 16) C (3, 4) D (16, 12)  Tangents are drawn to the circle x2 + y2 = 16 at the points Q where it intersects the circle x2 + y2 - 6x - 8y - 8 = 0, then the point of intersection of these tangents is A (4, 16/3) B (12, 16) C (3, 4) D (16, 12)  Tangents are drawn to the circle x2 + y2 = 16 at the points Q where it intersects the circle x2 + y2 - 6x - 8y - 8 = 0, then the point of intersection of these tangents is Solution: A C2 C1 P(x1 , y1) B Q Tangents are drawn to a unit circle with centre at origin from every point on the line 2x + y = 4. Find equation of locus of midpoint of CoCs. Solution: Orthogonality of two Circles  Orthogonality of two circles Two circles are said to be orthogonal if the tangents of the two circles at their point of intersection are perpendicular to each other.  Orthogonality of two circles Condition for Orthogonality r1 r2 C1 d C2 Two circles intersect each other orthogonally if r12 + r22 = d2 or 2g1g2 + 2f1f2 = c1 + c2  IIT 2000 If x2 + y2 + 2x + 2ky + 6 = 0 and x2 + y2 + 2ky + k = 0 Q intersect orthogonally then the possible values of k are ___. A B C D  IIT 2000 If x2 + y2 + 2x + 2ky + 6 = 0 and x2 + y2 + 2ky + k = 0 Q intersect orthogonally then the possible values of k are ___. A B C D  IIT 2000 If x2 + y2 + 2x + 2ky + 6 = 0 and x2 + y2 + 2ky + k = 0 Q intersect orthogonally then the possible values of k are ___. Solution:  The locus of centres of family of circle passing Q through the origin and cutting the circle x2 + y2 + 4x - 6y - 13 = 0 orthogonally, is A 4x + 6y + 13 = 0 B 4x - 6y + 13 = 0 C 4x + 6y - 13 = 0 D 4x - 6y - 13 = 0  The locus of centres of family of circle passing Q through the origin and cutting the circle x2 + y2 + 4x - 6y - 13 = 0 orthogonally, is A 4x + 6y + 13 = 0 B 4x - 6y + 13 = 0 C 4x + 6y - 13 = 0 D 4x - 6y - 13 = 0  The locus of centres of family of circle passing Q through the origin and cutting the circle x2 + y2 + 4x - 6y - 13 = 0 orthogonally, is Solution: Let the family of circles passing through origin be x2 + y2 + 2gx + 2fy = 0 They intersect circle x2 + y2 + 4x - 6y - 13 = 0 orthogonally. So, 2g(2) - 2f(3) = -13 Hence, locus of (-g, -f) is -4x + 6y + 13 = 0 ⇒ 4x - 6y - 13 = 0 Radical Axis and Radical Centre  Radical axis and Radical centre Radical axis Radical axis of two circles S = 0 and S’ = 0 is the locus of point whose powers with respect to the two given circles are equal. Its equation is given by S - S’ = 0. S=0 S’ = 0 S - S’ = 0 Radical axis and Radical centre Observations (1) When two circles are intersecting, then the radical axis is nothing but the common chord. (2) When two circles are touching each other, the radical axis is the common tangent at their point of contact. (3) Radical axis is always perpendicular to the line joining the two centres. (It also bisects the line segment joining the two centres if the circles are of equal radii) Q Find the equation of the radical axis of the circles 2x2 + 2y2 + 3x + 6y − 5 = 0 and 3x2 + 3y2 −7x + 8y − 11 = 0. Q Find the equation of the radical axis of the circles 2x2 + 2y2 + 3x + 6y − 5 = 0 and 3x2 + 3y2 −7x + 8y − 11 = 0. Solution: Q Find the equation of the radical axis of the circles 2x2 + 2y2 + 3x + 6y − 5 = 0 and 3x2 + 3y2 −7x + 8y − 11 = 0. Solution: Radical axis and Radical centre NOTE (1) Concentric circles do not have a radical axis.  Radical axis and Radical centre NOTE (1) Concentric circles do not have a radical axis. (2) Radical axes of 3 circles taken in pairs are concurrent. S=0 S − S’ = 0 S”− S = 0 Clearly, the lengths of the O tangents from O to all the S’ = 0 circles are equal. S” = 0 S’ - S” = 0  Radical axis and Radical centre Radical Centre Point of intersection of radical axes of 3 circles taken in pairs is called the radical centre of the circles. Radical axis and Radical centre Result (1) A circle with centre at the radical centre of three circles and radius equal to length of tangent (from radical centre) is orthogonal to all three circles. (2) The radical centre of three circles described on the side of a triangle as diameters is the orthocentre of the triangle.  The equation of the three circles are given : x2 + y2 = 1, Q x2 + y2 - 8x + 15 = 0 x2 + y2 + 10y + 24 = 0. Determine the coordinates of the point P such that the tangents drawn from it to the circles are equal in length. Solution: We know that the point from which lengths of tangents are equal in length is radical centre of the given three circles. Now radical axis of the first two circles is (x2 + y2 - 1) - (x2 + y2 - 8x + 15) = 0, i.e., x - 2 = 0, …(i) And radical axis of the second and third circles is (x2 + y2 - 8x + 15) - (x2 + y2 + 10y + 24) = 0, i.e., 8x + 10y + 9 = 0 …(ii) Solving Eqs. (i) and (ii), the coordinates of the radical centre, i.e., of point P are  The centre of the circle, which cuts each of the three Q circles: x2 + y2 + 2x + 17y + 4 = 0, x2 + y2 + 7x + 6y + 11 = 0 and x2 + y2 - x + 22y + 3 = 0, orthogonally is A (3, 2) B (1, 2) C (2, 3) D (0, 2)  The centre of the circle, which cuts each of the three Q circles: x2 + y2 + 2x + 17y + 4 = 0, x2 + y2 + 7x + 6y + 11 = 0 and x2 + y2 - x + 22y + 3 = 0, orthogonally is A (3, 2) B (1, 2) C (2, 3) D (0, 2) Solution: Let general equation of a circle is x2 + y2 + 2gx + 2fy + c = 0 …(i) If the circle (i) cuts orthogonally each of the given three circles. Then, condition is 2g1g2 + 2f1f2 = c1 + c2 Applying the condition one by one, we get 2g + 17f = c + 4 …(ii) 7g + 6f = c + 11 …(iii) -g + 22f = c + 3 …(iv) On solving Eqs. (ii), (iii) and (iv), we get g = -3, f = -2 Therefore, the centre of the circle is (3, 2). Alternate Solution S1 : x2 + y2 + 2x + 17y + 4 = 0 S2 : x2 + y2 + 7x + 6y + 11 = 0 S3 : x2 + y2 - x + 22y + 3 = 0 Radical axis for S1 and S2 S2 - S1 = 0 ⇒ 5x - 11y + 7 = 0 …(1) Radical axis for S2 and S3 S3 - S2 = 0 ⇒ x - 2y + 1 = 0 …(2) From (1) and (2) radical centre of the given circle will be (3, 2)  Find the radical centre of 3 circles described on the Q sides 4x − 7y + 10 = 0, x + y − 5 = 0 and 7x + 4y − 15 = 0 of a triangle, as diameters. A (1, 2) B (2, 1) C (1, -2) D (2, -1)  Find the radical centre of 3 circles described on the Q sides 4x − 7y + 10 = 0, x + y − 5 = 0 and 7x + 4y − 15 = 0 of a triangle, as diameters. A (1, 2) B (2, 1) C (1, -2) D (2, -1) Solution:  Unacademy Plus & Iconic Top Educators 1M+ Doubts with Video Solutions Interactive 100+ Daily Live Classes Access 1.3L+ hours of Recorded classes 250+ Test Series & 35K+Practice Section Compete Live 1:1 Mentorship Program (Iconic) Printed Notes (Iconic) Fill The Google Form available in Description https://docs.google.com/forms/d/1atjRl0T6zDkK- 6YnL5J5Us_5MaM8U8DZYn2B4TIWdrA/edit