Circles PDF - Past Paper

www.sakshieducation.com CIRCLES OBJECTIVES 1. If a circle passes through the point (0, 0), (a, 0), (0, b), then its centre is (a) (a, b)...

www.sakshieducation.com CIRCLES OBJECTIVES 1. If a circle passes through the point (0, 0), (a, 0), (0, b), then its centre is (a) (a, b) (b) (b, a) a b (c)  ,  (d)  b , − a  2 2 2 2 2. If one end of a diameter of the circle x 2 + y 2 − 4 x − 6 y + 11 = 0 be (3, 4), then the other end is (a) (0, 0) (b) (1, 1) (c) (1, 2) (d) (2, 1) 3. The equation of the circle passing through the origin and cutting intercepts of length 3 and 4 units from the positive axes, is (a) x 2 + y 2 + 6x + 8y + 1 = 0 (b) x 2 + y 2 − 6 x − 8 y = 0 (c) x 2 + y 2 + 3x + 4y = 0 (d) x 2 + y 2 − 3 x − 4 y = 0 4. If the vertices of a triangle be (2, − 2) , (−1, − 1) and (5, 2), then the equation of its circumcircle is (a) x 2 + y 2 + 3 x + 3y + 8 = 0 (b) x 2 + y 2 − 3 x − 3 y − 8 = 0 (c) x 2 + y 2 − 3 x + 3y + 8 = 0 (d) None of these 5. The equation of the circle having centre (1, − 2) and passing through the point of intersection of lines 3 x + y = 14 , 2 x + 5 y = 18 is (a) x 2 + y 2 − 2 x + 4 y − 20 = 0 (b) x 2 + y 2 − 2 x − 4 y − 20 = 0 (c) x 2 + y 2 + 2 x − 4 y − 20 = 0 (d) x 2 + y 2 + 2 x + 4 y − 20 = 0 www.sakshieducation.com  www.sakshieducation.com 6. For all values of θ , the locus of the point of intersection of the lines x cos θ + y sin θ = a and x sin θ − y cos θ = b is (a) An ellipse (b) A circle (c) A parabola (d) A hyperbola 7. The lines 2 x − 3y = 5 and 3x − 4y = 7 are the diameters of a circle of area 154 square units. The equation of the circle is (a) x 2 + y 2 + 2 x − 2 y = 62 (b) x 2 + y 2 − 2 x + 2 y = 47 (c) x 2 + y 2 + 2 x − 2 y = 47 (d) x 2 + y 2 − 2 x + 2 y = 62 8. The locus of the centre of the circle which cuts a chord of length 2a from the positive x- axis and passes through a point on positive y-axis distant b from the origin is (a) x 2 + 2by = b 2 + a 2 (b) x 2 − 2by = b 2 + a 2 (c) x 2 + 2by = a 2 − b 2 (d) x 2 − 2by = b 2 − a 2 9. The locus of the centre of the circle which cuts off intercepts of length 2a and 2b from x- axis and y-axis respectively, is (a) x +y = a+b (b) x 2 + y 2 = a2 + b 2 (c) x 2 − y 2 = a2 − b 2 (d) x 2 + y 2 = a2 − b 2 10. A circle touches x-axis and cuts off a chord of length 2l from y-axis. The locus of the centre of the circle is (a) A straight line (b) A circle (c) An ellipse (d) A hyperbola 11. A square is inscribed in the circle x 2 + y 2 − 2 x + 4 y + 3 = 0 , whose sides are parallel to the coordinate axes. One vertex of the square is (a) (1 + 2 , − 2) (b) (1 − 2 , − 2) (c) (1, − 2 + 2 ) (d) None of these 12. The equation of the circle whose radius is 5 and which touches the circle x 2 + y 2 − 2 x − 4 y − 20 = 0 externally at the point (5, 5), is (a) x 2 + y 2 − 18 x − 16 y − 120 = 0 (b) x 2 + y 2 − 18 x − 16 y + 120 = 0 (c) x 2 + y 2 + 18 x + 16 y − 120 = 0 (d) x 2 + y 2 + 18 x − 16 y + 120 = 0 www.sakshieducation.com  www.sakshieducation.com 13. The circle represented by the equation x 2 + y 2 + 2 gx + 2 fy + c = 0 will be a point circle, if (a) g2 + f 2 = c (b) g 2 + f 2 >c (c) g2 + f 2 + c = 0 (d) None of these 14. The number of circle having radius 5 and passing through the points (– 2, 0) and (4, 0) is (a) One (b) Two (c) Four (d) Infinite 15. The area of the circle whose centre is at (1, 2) and which passes through the point (4, 6) is (a) 5π (b) 10π (c) 25 π (d) None of these 16. The equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y − 4x + 3 = 0 , is (a) x 2 + y 2 + 4 x − 10 y + 25 = 0 (b) x 2 + y 2 − 4 x − 10 y + 25 = 0 (c) x 2 + y 2 − 4 x − 10 y + 16 = 0 (d) x 2 + y 2 − 14 y + 8 = 0 17. The equation of the circle touching x = 0, y = 0 and x =4 is (a) x 2 + y 2 − 4 x − 4 y + 16 = 0 (b) x 2 + y 2 − 8 x − 8 y + 16 = 0 (c) x 2 + y 2 + 4 x + 4y + 4 = 0 (d) x 2 + y 2 − 4 x − 4y + 4 = 0 18. The equation of the circle passing through the points (0, 0), (0, b) and (a, b) is (a) x 2 + y 2 + ax + by = 0 (b) x 2 + y 2 − ax + by = 0 (c) x 2 + y 2 − ax − by = 0 (d) x 2 + y 2 + ax − by = 0 19. If the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 touches x-axis, then (a) g= f (b) g 2 =c (c) f2 = c (d) g 2 + f 2 =c 20. The equation ax 2 + by 2 + 2hxy + 2 gx + 2 fy + c = 0 will represent a circle, if (a) a = b = 0 and c=0 (b) f =g and h=0 (c) a=b ≠0 and h=0 (d) f =g and c=0 21. A circle is concentric with the circle x 2 + y 2 − 6 x + 12 y + 15 = 0 and has area double of its area. The equation of the circle is (a) x 2 + y 2 − 6 x + 12 y − 15 = 0 (b) x 2 + y 2 − 6 x + 12 y + 15 = 0 (c) x 2 + y 2 − 6 x + 12 y + 45 = 0 (d)None of these www.sakshieducation.com  www.sakshieducation.com 22. The equation of the circle with centre at (1, –2) and passing through the centre of the given circle x 2 + y 2 + 2y − 3 = 0 , is (a) x 2 + y 2 − 2x + 4 y + 3 = 0 (b) x 2 + y 2 − 2 x + 4 y − 3 = 0 (c) x 2 + y 2 + 2x − 4 y − 3 = 0 (d) x 2 + y 2 + 2 x − 4 y + 3 = 0 23. If the radius of the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 be r, then it will touch both the axes, if (a) g= f =r (b) g = f =c=r (c) g= f = c =r (d) g = f and c2 = r 24. If the lines x+y =6 and x + 2y = 4 be diameters of the circle whose diameter is 20, then the equation of the circle is (a) x 2 + y 2 − 16 x + 4 y − 32 = 0 (b) x 2 + y 2 + 16 x + 4 y − 32 = 0 (c) x 2 + y 2 + 16 x + 4 y + 32 = 0 (d) x 2 + y 2 + 16 x − 4 y + 32 = 0 25. For the circle x 2 + y 2 + 3 x + 3y = 0 , which of the following relations is true (a) Centre lies on x-axis (b)Centre lies on y-axis (c) Centre is at origin (d)Circle passes through origin 26. The equation of the circle passing through the point (2, 1) and touching y-axis at the origin is (a) x 2 + y 2 − 5x = 0 (b) 2 x 2 + 2 y 2 − 5 x = 0 (c) x 2 + y 2 + 5x = 0 (d) None of these 27. Equation of a circle whose centre is origin and radius is equal to the distance between the lines x =1 and x = −1 is (a) x2 + y2 = 1 (b) x 2 + y 2 = 2 (c) x2 + y2 = 4 (d) x 2 + y 2 = −4 28. The centre and radius of the circle 2 x 2 + 2y 2 − x = 0 are (a)  1 , 0  and 1 (b)  − 1 , 0  and 1 4  4  2  2 1  (c)  , 0 and 1 (d)  0, − 1  and 1 2  2  4 4 www.sakshieducation.com  www.sakshieducation.com 29. The equation of the circle which touches both the axes and whose radius is a, is (a) x 2 + y 2 − 2 ax − 2 ay + a 2 = 0 (b) x 2 + y 2 + ax + ay − a 2 =0 (c) x 2 + y 2 + 2 ax + 2 ay − a 2 = 0 (d) x 2 + y 2 − ax − ay + a 2 =0 30. A circle which passes through origin and cuts intercepts on axes a and b, the equation of circle is (a) x 2 + y 2 − ax − by = 0 (b) x 2 + y 2 + ax + by = 0 (c) x 2 + y 2 − ax + by = 0 (d) x 2 + y 2 + ax − by = 0 31. The radius of a circle which touches y-axis at (0,3) and cuts intercept of 8 units with x- axis, is (a) 3 (b) 2 (c) 5 (d) 8 32. The equation of the circumcircle of the triangle formed by the lines y + 3 x = 6, y − 3 x = 6 , and y=0 , is (a) x 2 + y 2 − 4y = 0 (b) x 2 + y 2 + 4 x = 0 (c) x 2 + y 2 − 4 y = 12 (d) x 2 + y 2 + 4 x = 12 33. If the lines l1 x + m 1 y + n1 = 0 and l2 x + m 2 y + n2 = 0 cuts the axes at con-cyclic points, then (a) l1 l 2 = m1m 2 (b) l1 m 1 = l2 m 2 (c) l1 l 2 + m 1 m 2 = 0 (d) l1 m 2 = l2 m 1 34. The equation of a circle with centre (−4 , 3) and touching the circle x2 + y2 = 1 , is (a) x 2 + y 2 + 8 x − 6y + 9 = 0 (b) x 2 + y 2 + 8 x + 6 y − 11 = 0 (c) x 2 + y 2 + 8 x + 6y − 9 = 0 (d)None of these 35. A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If m and n are the distance of the tangents to the circle at the points A and B respectively from the origin, the diameter of the circle is (a) m (m + n) (b) m + n 1 (c) n(m + n) (d) (m + n) 2 www.sakshieducation.com  www.sakshieducation.com 36. The equation to a circle whose centre lies at the point (–2, 1) and which touches the line 3 x − 2y − 6 = 0 at (4, 3), is (a) x 2 + y 2 + 4 x − 2 y − 35 = 0 (b) x 2 + y 2 − 4 x + 2 y + 35 = 0 (c) x 2 + y 2 + 4 x + 2 y + 35 = 0 (d)None of these 37. The equation of circle whose diameter is the line joining the points (–4, 3) and (12, –1) is (a) x 2 + y 2 + 8 x + 2 y + 51 = 0 (b) x 2 + y 2 + 8 x − 2 y − 51 = 0 (c) x 2 + y 2 + 8 x + 2 y − 51 = 0 (d) x 2 + y 2 − 8 x − 2 y − 51 = 0 38. If (α , β ) is the centre of a circle passing through the origin, then its equation is (a) x 2 + y 2 − αx − βy = 0 (b) x 2 + y 2 + 2αx + 2 βy = 0 (c) x 2 + y 2 − 2αx − 2 βy = 0 (d) x 2 + y 2 + αx + βy = 0 39. The locus of the centre of a circle which touches externally the circle x 2 + y 2 − 6 x − 6 y + 14 = 0 and also touches the y-axis, is given by the equation (a) x 2 − 6 x − 10 y + 14 = 0 (b) x 2 − 10 x − 6 y + 14 =0 (c) y 2 − 6 x − 10 y + 14 = 0 (d) y 2 − 10 x − 6 y + 14 =0 π 40. Area of the circle in which a chord of length 2 makes an angle at the centre is 2 π (a) (b) 2π 2 (c) π (d) π 4 41. For ax 2 + 2 hxy + 3 y 2 + 4 x + 8 y − 6 = 0 to represent a circle, one must have (a) a = 3, h = 0 (b) a = 1, h = 0 (c) a=h=3 (d) a = h = 0 42. Circles are drawn through the point (2, 0) to cut intercept of length 5 units on the x-axis. If their centres lie in the first quadrant, then their equation is (a) x 2 + y 2 + 9 x + 2 fy + 14 = 0 (b) 3 x 2 + 3 y 2 + 27 x − 2 fy + 42 = 0 (c) x 2 + y 2 − 9 x + 2 fy + 14 = 0 (d) x 2 + y 2 − 2 fy − 9 y + 14 = 0 43. Equations to the circles which touch the lines 3 x − 4 y + 1 = 0 , 4 x + 3y − 7 = 0 and pass through (2, 3) are (a) (x − 2)2 + (y − 8 ) 2 = 25 (b) 5 x 2 + 5 y 2 − 12 x − 24 y + 31 = 0 (c) Both (a) and (b) (d) None of these www.sakshieducation.com  www.sakshieducation.com 44. The equation of the circle in the first quadrant which touches each axis at a distance 5 from the origin is (a) x 2 + y 2 + 5 x + 5 y + 25 = 0 (b) x 2 + y 2 − 10 x − 10 y + 25 = 0 (c) x 2 + y 2 − 5 x − 5 y + 25 = 0 (d) x 2 + y 2 + 10 x + 10 y + 25 = 0 45. The equation of circle whose centre lies on 3 x − y − 4 = 0 and x + 3 y + 2 = 0 and has an area 154 square units is (a) x 2 + y 2 − 2 x + 2 y − 47 = 0 (b) x 2 + y 2 − 2 x + 2 y + 47 = 0 (c) x 2 + y 2 + 2 x − 2 y − 47 = 0 (d) None of these 46. The equation of circle with centre (1, 2) and tangent x +y −5 = 0 is (a) x 2 + y 2 + 2x − 4 y + 6 = 0 (b) x 2 + y 2 − 2x − 4y + 3 = 0 (c) x 2 + y 2 − 2x + 4 y + 8 = 0 (d) x 2 + y 2 − 2x − 4y + 8 = 0 47. The equation of the circle of radius 5 and touching the coordinate axes in third quadrant is (a) (x − 5 ) 2 + (y + 5 ) 2 = 25 (b) (x + 4 ) 2 + (y + 4 ) 2 = 25 (c) (x + 6 ) 2 + (y + 6 ) 2 = 25 (d) (x + 5) 2 + (y + 5) 2 = 25 48. If the lines 2 x + 3y + 1 = 0 and 3 x − y − 4 = 0 lie along diameters of a circle of circumference 10π , then the equation of the circle is (a) x 2 + y 2 + 2 x − 2 y − 23 = 0 (b) x 2 + y 2 − 2 x − 2 y − 23 = 0 (c) x 2 + y 2 + 2 x + 2 y − 23 = 0 (d) x 2 + y 2 − 2 x + 2 y − 23 = 0 49. For what value of k, the points (0, 0), (1, 3), (2, 4) and (k, 3) are con-cyclic (a) 2 (b) 1 (c) 4 (d) 5 50. If g2 + f 2 = c , then the equation x 2 + y 2 + 2 gx + 2 fy + c = 0 will represent (a) A circle of radius g (b)A circle of radius f (c) A circle of diameter c (d) A circle of radius 0 51. A variable circle passes through the fixed point (2,0) and touches the y-axis. Then the locus of its centre is (a) A circle (b) An Ellipse (c) A hyperbola (d) A parabola www.sakshieducation.com  www.sakshieducation.com 52. The length of intercept, the circle x 2 + y 2 + 10 x − 6 y + 9 = 0 makes on the x-axis is (a) 2 (b) 4 (c) 6 (d) 8 53. The centre of the circle x = 2 + 3 cos θ , y = 3 sin θ − 1 is (a) (3, 3) (b) (2, − 1) (c) (−2, 1) (d) (−1, 2) 54. The four distinct points (0, 0),(2, 0), (0, –2) and (k,–2)are con-cyclic, if k = (a) –2 (b) 2 (c) 1 (d) 0 55. Let P ( x 1 , y1 ) and Q(x 2 , y 2 ) are two points such that their abscissa x1 and x2 are the roots of the equation x 2 + 2x − 3 = 0 while the ordinates y1 and y2 are the roots of the equation y 2 + 4 y − 12 = 0. The centre of the circle with PQ as diameter is (a) (−1,−2) (b) (1, 2) (c) (1,−2) (d) (−1,2) 56. If one end of the diameter is (1, 1) and other end lies on the line x +y =3, then locus of centre of circle is (a) x + y = 1 (b) 2( x − y ) = 5 (c) 2 x + 2y = 5 (d) None of these 57. A circle is drawn to cut a chord of length 2a units along X-axis and to touch the Y-axis. The locus of the centre of the circle is (a) x 2 + y 2 = a 2 (b) x 2 − y 2 = a 2 (c) x + y = a 2 (d) x 2 − y 2 = 4 a 2 (e) x 2 + y 2 = 4 a2 58. If the length of tangent drawn from the point (5, 3) to the circle x 2 + y 2 + 2 x + ky + 17 = 0 be 7, then k = (a) 4 (b) – 4 (c) – 6 (d) 13/2 www.sakshieducation.com  www.sakshieducation.com 59. If OA and OB be the tangents to the circle x 2 + y 2 − 6 x − 8 y + 21 = 0 drawn from the origin O, then AB = 4 (a) 11 (b) 21 5 17 (c) (d) None of these 3 60. The equations of the tangents to the circle x 2 + y 2 = 50 at the points where the line x +7 = 0 meets it, are (a) 7 x ± y + 50 = 0 (b) 7 x ± y − 5 = 0 (c) y ± 7x + 5 = 0 (d) y ± 7 x − 5 = 0 61. The line ( x − a) cos α + (y − b ) sin α = r will be a tangent to the circle (x − a) 2 + (y − b) 2 = r 2 (a) If α = 30 o (b) If α = 60 o (c) For all values of α (d) None of these  1 1  62. The equation of the normal to the circle x2 + y2 = 9 at the point   2 , 2  is   2 (a) x+y =0 (b) x − y = 3 (c) x−y =0 (d) None of these 63. The equations of the tangents drawn from the point (0, 1) to the circle x 2 + y 2 − 2x + 4y = 0 are (a) 2 x − y + 1 = 0, x + 2 y − 2 = 0 (b) 2 x − y + 1 = 0, x + 2 y + 2 = 0 (c) 2 x − y − 1 = 0, x + 2 y − 2 = 0 (d) 2 x − y − 1 = 0, x + 2 y + 2 = 0 64. The equations of the tangents to the circle x 2 + y 2 = 36 which are inclined at an angle of 45 o to the x-axis are (a) x+y =± 6 (b) x = y ± 3 2 (c) y = x ±6 2 (d) None of these 65. The length of tangent from the point (5, 1) to the circle x 2 + y 2 + 6x − 4y − 3 = 0 , is (a) 81 (b) 29 (c) 7 (d) 21 66. If the line lx + my = 1 be a tangent to the circle x 2 + y 2 = a2 , then the locus of the point (l, m) is (a) A straight line (b) A Circle (c) A parabola (d) An ellipse www.sakshieducation.com  www.sakshieducation.com 67. The equations of the normals to the circle x 2 + y 2 − 8 x − 2 y + 12 = 0 at the points whose ordinate is –1, will be (a) 2 x − y − 7 = 0, 2 x + y − 9 = 0 (b) 2 x + y + 7 = 0, 2 x + y + 9 = 0 (c) 2 x + y − 7 = 0, 2 x + y + 9 = 0 (d) 2 x − y + 7 = 0, 2 x − y + 9 = 0 68. If the line x = k touches the circle x2 + y2 = 9 , then the value of k is (a) 2 but not – 2 (b) – 2 but not 2 (c) 3 (d) None of these 69. If the ratio of the lengths of tangents drawn from the point ( f , g ) to the given circle x 2 + y 2 = 6 and x 2 + y 2 + 3 x + 3 y = 0 be 2 : 1, then (a) f 2 + g2 + 2g + 2 f + 2 = 0 (b) f 2 + g2 + 4 g + 4 f + 4 = 0 (c) f 2 + g2 + 4 g + 4 f + 2 = 0 (d) None of these 70. The line y = mx + c will be a normal to the circle with radius r and centre at (a, b), if (a) a = mb + c (b) b = ma + c (c) r = ma − b + c (d) r = ma − b  ab 2 a 2b  a 2b 2 71. The equation of the tangent at the point  ,  of the circle x 2 + y2 = is  a2 + b 2 a2 + b 2  a2 + b 2   x y x y (a) + =1 (b) + +1 = 0 a b a b x y x y (c) − =1 (d) − +1 = 0 a b a b 72. Two tangents drawn from the origin to the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 will be perpendicular to each other, if (a) g 2 + f 2 = 2c (b) g = f = c 2 (c) g+ f =c (d) None of these x y 73. The equation of circle which touches the axes of coordinates and the line + = 1 and 3 4 whose centre lies in the first quadrant is x 2 + y 2 − 2 cx − 2 cy + c 2 = 0 , where c is (a) 1 (b) 2 (c) 3 (d) 6 www.sakshieducation.com  www.sakshieducation.com 74. A tangent to the circle x 2 + y 2 = 5 at the point (1,–2)..... the circle x 2 + y 2 − 8 x + 6 y + 20 = 0 (a) Touches (b) Cuts at real points (c) Cuts at imaginary points (d) None of these 75. Square of the length of the tangent drawn from the point (α , β ) to the circle ax 2 + ay 2 = r 2 is 2 (a) aα 2 + aβ 2 − r 2 (b) α 2 + β 2 − r a r2 (c) α2 + β 2 + (d) α 2 + β 2 − r 2 a 76. The number of common tangents to the circles x 2 + y 2 − 4 x − 6 y − 12 = 0 and x 2 + y 2 + 6 x + 18 y + 26 = 0 is (a) 1 (b) 2 (c) 3 (d) 4 77. The area of triangle formed by the tangent, normal drawn at (1, 3 ) to the circle x 2 + y2 = 4 and positive x-axis, is (a) 2 3 (b) 3 (c) 4 3 (d) None of these 78. Line y = x +a 2 is a tangent to the circle x 2 + y 2 = a 2 at (a)  a , a   (b)  − a ,− a    2 2   2 2   a a   a a  (c)  ,−  (d)  − ,   2 2   2 2  79. Length of the tangent from (x 1 , y1 ) to the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 is (a) (x 12 + y12 + 2 gx 1 + 2 fy1 + c)1 / 2 (b) (x 12 + y12 )1 / 2 (c) [(x 1 + g)2 + (y1 + f )2 ]1 / 2 (d) None of these 80. The points of intersection of the line 4 x − 3 y − 10 = 0 and the circle x 2 + y 2 − 2 x + 4 y − 20 = 0 are (a) (−2,−6 ), (4 ,2) (b) (2, 6), (−4,−2) (c) (−2, 6), (−4 , 2) (d) None of these 81. The angle between the tangents from (α , β ) to the circle x 2 + y 2 = a2 , is    α 2 + β 2 − a2  (a) tan −1  a   (b) tan − 1     α 2 + β 2 − a2  a        2 tan −1   a (c)  (d) None of these  α + β 2 − a2 2   www.sakshieducation.com  www.sakshieducation.com 82. The gradient of the tangent line at the point (a cos α , a sin α ) to the circle x 2 + y 2 = a2 , is (a) tan α (b) tan( π − α ) (c) cot α (d) − cot α 83. The line y = mx + c intersects the circle x 2 + y 2 = r 2 at two real distinct points, if (a) −r 1 + m2 < c ≤ 0 (b) 0 ≤ c < r 1 + m2 (c) (a) and (b) both (d) − c 1 −m2 < r 84. 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, the area of the quadrilateral OACB is 1 (a) c (g 2 + f 2 − c ) (b) c( g 2 + f 2 − c ) 2 g2 + f 2 − c (c) c g2 + f 2 − c (d) c 85. If a circle passes through the points of intersection of the coordinate axis with the lines λx − y + 1 = 0 and x − 2 y + 3 = 0 , then the value of λ is (a) 1 (b) 2 (c) 3 (d) 4 86. The equation of the tangent to the circle x 2 + y2 − 2x − 4y − 4 = 0 which is perpendicular to 3x − 4y −1 = 0 , is (a) 4 x + 3y − 5 = 0 (b) 4 x + 3 y + 25 = 0 (c) 4 x − 3y + 5 = 0 (d) 4 x + 3 y − 25 = 0 87. Given the circles x 2 + y2 − 4 x − 5 = 0 and x 2 + y 2 + 6 x − 2y + 6 = 0. Let P be a point (α , β ) such that the tangents from P to both the circles are equal, then (a) 2α + 10 β + 11 = 0 (b) 2α − 10 β + 11 = 0 (c) 10 α − 2 β + 11 = 0 (d) 10α + 2 β + 11 = 0 88. If a > 2b > 0 then the positive value of m for which y = mx − b 1 + m 2 is a common tangent to x 2 + y2 = b2 and (x − a)2 + y 2 = b 2 , is 2b a2 − 4 b 2 (a) (b) a2 − 4 b 2 2b 2b b (c) (d) a − 2b a − 2b www.sakshieducation.com  www.sakshieducation.com 89. 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 (a)  −7 ,−4  (b)  −18 , −21   2   5 5  (c)(2,–7) (d) (–2, –5) 90. The tangent at P, any point on the circle x 2 + y2 = 4 , meets the coordinate axes in A and B, then (a) Length of AB is constant (b) PA and PB are always equal (c) The locus of the mid point of AB is x 2 + y 2 = x 2y 2 (d) None of these 91. If the circle (x − h)2 + (y − k )2 = r 2 touches the curve y = x 2 + 1 at a point (1, 2), then the possible locations of the points (h, k) are given by (a) hk = 5 / 2 (b) h + 2k = 5 (c) h2 − 4 k 2 = 5 (d) k 2 = h 2 + 1 92. The line ax + by + c = 0 is a normal to the circle x 2 + y 2 = r2. The portion of the line ax + by + c = 0 intercepted by this circle is of length (a) r (b) r 2 (c) 2r (d) r 93. The gradient of the normal at the point (–2, –3) on the circle x 2 + y 2 + 2 x + 4 y + 3 = 0 is (a) 1 (b) –1 3 1 (c) (d) 2 2 94. A circle with centre (a, b) passes through the origin. The equation of the tangent to the circle at the origin is (a) ax − by = 0 (b) ax + by = 0 (c) bx − ay = 0 (d) bx + ay = 0 x y 95. If + =1 touches the circle x 2 + y 2 = a2 , then point (1 / α , 1 / β ) lies on a/an α β (a)Straight line (b) Circle (c) Parabola (d) Ellipse www.sakshieducation.com  www.sakshieducation.com 96. Assertion (a) : The circle x 2 + y 2 = 1 has exactly two tangents parallel to the x-axis dy Reason (R) : =0 on the circle exactly at the point (0,±1). Of these statements dx (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true 97. If 5 x − 12 y + 10 = 0 and 12 y − 5 x + 16 = 0 are two tangents to a circle, then the radius of the circle is (a) 1 (b) 2 (c) 4 (d) 6 98. The square of the length of the tangent from (3, –4) on the circle x 2 + y 2 − 4 x − 6 y + 3 = 0 is (a) 20 (b) 30 (c) 40 (d) 50 99. The locus of a point which moves so that the ratio of the length of the tangents to the circles x 2 + y2 + 4 x + 3 = 0 and x 2 + y2 − 6x + 5 = 0 is 2:3 is (a) 5 x 2 + 5 y 2 − 60 x + 7 = 0 (b) 5 x 2 + 5 y 2 + 60 x − 7 = 0 (c) 5 x 2 + 5 y 2 − 60 x − 7 = 0 (d) 5 x 2 + 5 y 2 + 60 x + 7 = 0 100. The distance between the chords of contact of the tangents to the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 from the origin and the point (g, f ) is  g2 + f 2 − c   g2 + f 2 − c  (a) 1   (b)    2    2   2  g 2 + f   g 2 + f  1  g2 + f 2 − c  (c)   (d) None of these 2  g2 + f 2    101. If the middle point of a chord of the circle x 2 + y 2 + x − y − 1 = 0 be (1, 1), then the length of the chord is (a) 4 (b) 2 (c) 5 (d) None of these www.sakshieducation.com  www.sakshieducation.com 102. Locus of the middle points of the chords of the circle x 2 + y 2 = a 2 which are parallel to y = 2x will be 1 (a) A circle with radius a (b) A straight line with slope − 2 (c) A circle will centre (0, 0) (d) A straight line with slope – 2 103. The equation of the common chord of the circles (x − a)2 + (y − b )2 = c 2 and (x − b )2 + (y − a)2 = c 2 is (a) x −y = 0 (b) x + y = 0 (c) x + y = a2 + b 2 (d) x − y = a 2 − b 2 104. The co-ordinates of pole of line lx + my + n = 0 with respect to circles x 2 + y2 = 1 , is (a)  l , m  (b)  − l ,− m  n n   n n l m (c)  ,−  (d)  − l , m  n n  n n x y 105. The length of the chord intercepted by the circle x 2 + y 2 = r 2 on the line + =1 is a b r 2 (a 2 + b 2 ) − a 2b 2 r 2 (a 2 + b 2 ) − a 2b 2 (a) (b) 2 a2 + b 2 a2 + b 2 r 2 (a 2 + b 2 ) − a 2 b 2 (c) 2 (d) None of these a2 + b 2 106. A line lx + my + n = 0 meets the circle x 2 + y 2 = a 2 at the points P and Q. The tangents drawn at the points P and Q meet at R, then the coordinates of R is     (b)  − a l , − a m  2 2 2 2 (a)  a l , a m   n n   n n   a 2n a 2 n  (c)   (d) None of these  l , m    107. The length of the common chord of the circles x 2 + y 2 + 2 x + 3y + 1 = 0 and x 2 + y 2 + 4 x + 3 y + 2 = 0 is (a) 9/2 (b) 2 2 (c) 3 2 (d) 3 / 2 108. Length of the common chord of the circles x 2 + y2 + 5 x + 7y + 9 = 0 and x 2 + y2 + 7x + 5y + 9 = 0 is (a) 9 (b) 8 (c) 7 (d)6 www.sakshieducation.com  www.sakshieducation.com 109. If polar of a circle x 2 + y 2 = a 2 with respect to (x ' , y ' ) is Ax + By + C = 0 , then its pole will be  A a 2 B  2  A a 2 B  2 (a)  a ,  (b)  a ,  −C −C   C C   a 2C a 2C   C a 2 C  2 (c)   A , B   (d)  a ,     −A −B  110. If the circle x 2 + y 2 = a 2 cuts off a chord of length 2b from the line y = mx + c , then (a) (1 − m 2 )(a 2 + b 2 ) = c 2 (b) (1 + m 2 )(a 2 − b 2 ) = c 2 (c) (1 − m 2 )(a 2 − b 2 ) = c 2 (d) None of these 111. The radius of the circle, having centre at (2,1) whose one of the chord is a diameter of the circle x 2 + y 2 − 2x − 6y + 6 = 0 is (a) 1 (b) 2 (c) 3 (d) 3 112. The intercept on the line y=x by the circle x 2 + y 2 − 2x = 0 is AB, equation of the circle on AB as a diameter is (a) x 2 + y2 + x − y = 0 (b) x 2 + y 2 − x + y = 0 (c) x 2 + y2 + x + y = 0 (d) x 2 + y 2 − x − y = 0 113. A line through (0,0) cuts the circle x 2 + y 2 − 2 ax = 0 at A and B, then locus of the centre of the circle drawn on AB as a diameter is (a) x 2 + y 2 − 2ay = 0 (b) x 2 + y 2 + ay = 0 (c) x 2 + y 2 + ax = 0 (d) x 2 + y 2 − ax = 0 114. From the origin chords are drawn to the circle ( x − 1)2 + y 2 = 1. The equation of the locus of the middle points of these chords is (a) x2 + y2 − 3x = 0 (b) x 2 + y 2 − 3 y = 0 (c) x 2 + y2 − x = 0 (d) x 2 + y 2 − y = 0 115. If the line x − 2y = k cuts off a chord of length 2 from the circle x2 + y2 = 3 , then k = (a) 0 (b) ±1 (c) ± 10 (d) None of these www.sakshieducation.com  www.sakshieducation.com 116. The equation of the chord of the circle x 2 + y 2 = a 2 having (x1 , y1 ) as its mid-point is (a) xy 1 + yx 1 = a 2 (b) x1 + y1 = a (c) xx 1 + yy 1 = x 12 + y12 (d) xx 1 + yy 1 = a 2 117. The equation of the circle with origin as centre passing the vertices of an equilateral triangle whose median is of length 3a is (a) x 2 + y 2 = 9a 2 (b) x 2 + y 2 = 16 a 2 (c) x 2 + y 2 = a2 (d) None of these  1  118. If  m i ,  , i = 1, 2,3,4 are con-cyclic points, then the value of m1.m 2.m 3.m 4 is  mi  (a) 1 (b) – 1 (c) 0 (d) None of these 119. Tangents are drawn from the point (4, 3) to the circle x2 + y2 = 9. The area of the triangle formed by them and the line joining their points of contact is 24 64 (a) (b) 25 25 192 (c) (d) 192 25 5 120. Let L1 be a straight line passing through the origin and L 2 be the straight line x +y =1. If the intercepts made by the circle x 2 + y 2 − x + 3 y = 0 on L1 and L 2 are equal, then which of the following equations can represent L1 (a) x+y =0 (b) x − y = 0 (c) x − 7 y = 0 (d) x − 7 y = 0 121. The two points A and B in a plane are such that for all points P lies on circle satisfied PA =k , then k will not be equal to PB (a) 0 (b) 1 (c) 2 (d) None of these 122. A circle is inscribed in an equilateral triangle of side a, the area of any square inscribed in the circle is a2 2a 2 2 2 (a) (b) (c) a (d) a 3 3 6 12 www.sakshieducation.com  www.sakshieducation.com 123. The area of the triangle formed by joining the origin to the points of intersection of the line x 5 + 2y = 3 5 and circle x 2 + y 2 = 10 is (a) 3 (b) 4 (c) 5 (d) 6 124. The abscissae of A and B are the roots of the equation x 2 + 2 ax − b 2 = 0 and their ordinates are the roots of the equation y 2 + 2 py − q 2 = 0. The equation of the circle with AB as diameter (a) x 2 + y 2 + 2 ax + 2 py − b 2 − q 2 = 0 (b) x 2 + y 2 + 2ax + py − b 2 − q 2 =0 (c) x 2 + y 2 + 2 ax + 2 py + b 2 + q 2 = 0 (d) None of these 125. Let PQ and RS be tangents at the extremeties 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 PQ + RS (a) PQ. RS (b) 2 2 PQ. RS PQ 2 + RS 2 (c) (d) PQ + RS 2 126. Let AB be a chord of the circle x 2 + y 2 = r2 subtending a right angle at the centre. Then the locus of the centroid of the ∆PAB as P moves on the circle is (a) A parabola (b) A circle (c) An ellipse (d) A pair of straight lines 127. If two distinct chords, drawn from the point (p, q) on the circle x 2 + y 2 = px + qy , (where pq ≠ 0 ) are bisected by the x-axis, then (a) p2 = q2 (b) p 2 = 8q 2 (c) p 2 < 8q 2 (d) p 2 > 8q 2 128. If a straight line through C (− 8 , 8 ) making an angle of 135 ° with the x-axis cuts the circle x = 5 cos θ , y = 5 sin θ at points A and B, then the length of AB is (a) 3 (b) 7 (c) 10 (d) None of these www.sakshieducation.com  www.sakshieducation.com 129. A chord AB drawn from the point A(0,3 ) on circle x 2 + 4 x + (y − 3)2 = 0 meets to M in such a way that AM = 2 AB , then the locus of point M will be (a) Straight line (b) Circle (c) Parabola (d) None of these 130. If a circle passes through the point (a, b)and cuts the circle x2 + y2 = 4 orthogonally, then locus of its centre is (a) 2ax − 2by − (a 2 + b 2 + 4 ) = 0 (b) 2ax + 2by − (a 2 + b 2 + 4 ) = 0 (c) 2ax − 2by + (a 2 + b 2 + 4 ) = 0 (d) 2ax + 2by + (a 2 + b 2 + 4 ) = 0 131. The locus of centre of the circle which touches the circle x 2 + (y − 1)2 = 1 externally and also touches x-axis is (a) {( x , y ) : x 2 + (y − 1)2 = 4 } ∪ {( x , y ) : y < 0} (b) {( x , y ) : x 2 = 4 y} ∪ {(0, y ) : y < 0} (c) {( x , y ) : x 2 = y } ∪ {(0, y ) : y < 0} (d) {( x , y ) : x 2 = 4 y} ∪ {( x , y ) : y < 0} 132. The tangents are drawn from the point (4, 5) to the circle x 2 + y 2 − 4 x − 2 y − 11 = 0. The area of quadrilateral formed by these tangents and radii, is (a) 15 sq. units (b) 75 sq. units (c)8 sq. units (d)4 sq. units www.sakshieducation.com  www.sakshieducation.com CIRCLES HINTS AND SOLUTIONS 1. (c) Let the equation of circle be x 2 + y 2 + 2 gx + 2 fy + c = 0. Now on passing through the points, we get three equations. c=0 ….(i) a 2 + 2 ga + c = 0 ….(ii) b 2 + 2 fb + c = 0 ….(iii) a b On solving them, we get g=− , f =− 2 2 a b Hence the centre is  , . 2 2 2. (c) Centre is (2, 3). One end is (3, 4). P2 divides the join of P1 and O in ratio of 2 : 1. 4 −3 6−4 Hence P2 is  ,  ≡ (1, 2).  2 −1 2 −1  3  3. (d) Obviously the centre of the circle is  , 2. 2  Therefore, the equation of circle is 2 2  3 5  x −  + (y − 2)2 =   ⇒ x 2 + y 2 − 3 x − 4 y = 0.  2 2 4. (b) Let us find the equation of family of circles through (2, − 2) and (−1, − 1).  y+2 x −2 i.e. ( x − 2)( x + 1) + (y + 2)(y + 1) + λ  − =0  − 2 +1 2 +1  Now for point (5, 2) to lie on it, we should have λ given by  4  30 3. 6 + 4. 3 + λ − 1 = 0 ⇒ λ = =6 −1  5 Hence equation is y +2 x −2 ( x − 2)( x + 1) + (y + 2)(y + 1) + 6  − =0  −1 3  Or x 2 + y 2 − 3 x − 3 y − 8 = 0. www.sakshieducation.com  www.sakshieducation.com 5. (a) The point of intersection of 3 x + y − 14 = 0 and 2 x + 5 y − 18 = 0 are −18 + 70 −28 + 54 x= ,y = ⇒ x = 4, y = 2 15 − 2 13 i.e., point is (4, 2). Therefore radius is (9) + (16 ) = 5 and equation is x 2 + y 2 − 2 x + 4 y − 20 = 0. 6. (b) The point of intersection is x = a cos θ + b sin θ y = a sin θ − b cos θ. Therefore, x 2 + y 2 = a2 + b 2. Obviously, it is equation of a circle. 7. (d) If the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 touches the x-axis, then −f = g2 + f 2 − c ⇒ g2 = c.....(i) and cuts a chord of length 2l from y-axis ⇒ 2 f 2 − c = 2l ⇒ f 2 − c = l 2 ….(ii) Subtracting (i) from (ii), we get f 2 − g2 = l2. Hence the locus is y 2 − x 2 = l2 , which is obviously a hyperbola. 8. (d) Obviously the centre of the given circle is (1,− 2). Since the sides of square are parallel to the axes, therefore, first three alternates cannot be vertices of square because in first two (a and b) y = −2 and in (c) x = 1 , which passes through centre (1, − 2) but it is not possible. Hence answer (d) is correct. 9. (b) Let the centre of the required circle be (x 1 , y 1 ) and the centre of given circle is (1, 2). Since radii of both circles are same, therefore, point of contact (5, 5) is the mid point of the line joining the centres of both circles. Hence x1 = 9 and y1 = 8. Hence the required equation is (x − 9 ) 2 + (y − 8 ) 2 = 25 10. (a) Using condition of point circle R= g2 + f2 − c = 0 ⇒ g2 + f 2 = c. 11. (b) Two, centre of each lying on the perpendicular bisector of the join of the two points. 12. (c) Obviously radius = (1 − 4 ) 2 + (2 − 6 ) 2 = 5 Hence the area is given by πr 2 = 25 π sq. units. 13. (b) First find the centre. Let centre be (h, k), then www.sakshieducation.com  www.sakshieducation.com (h − 2) 2 + (k − 3) 2 = (h − 4 ) 2 + (k − 5 ) 2 ….(i) and k − 4h + 3 = 0 ….(ii) From (i), we get −4 h − 6 k + 8 h + 10 k = 16 + 25 − 4 − 9 Or 4 h + 4 k − 28 = 0 or h +k −7 = 0 ….(iii) From (iii) and (ii), we get (h, k) as (2, 5). Hence centre is (2, 5) and radius is 2. Now find the equation of circle. 14. (d) (x − 2) 2 + (y − 2) 2 = 4 x 2 + 4 − 4 x + y2 + 4 − 4y = 4 15. (c) Equation of circle passing through (0, 0) is x 2 + y 2 + 2 gx + 2 fy = 0 ….(i) Also, circle (i) is passing through (0, b) and (a, b) b  b ∴ f =− and a 2 + b 2 + 2 ag + 2  −  b = 0 2  2 a ⇒g=− 2 Hence the equations of circle is, x 2 + y 2 − ax − by = 0. 16. (b) Touches x-axis, hence radius = ordinate of centre. Hence g 2 + f 2 − c = (− f ) or g2 = c. 17. (b) Centre of circle = Point of intersection of diameters = (1, –1) Now area = 154 ⇒ πr 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. 18. (c) Here 2 g 2 − c = 2a ⇒ g 2 − a 2 − c = 0.....(i) and it passes through (0, b), therefore b 2 + 2 fb + c = 0 ….(ii) On adding (i) and (ii), we get g 2 + 2 fb = a 2 − b 2 Hence locus is x 2 + 2by = a 2 − b 2. 19. (c) 2 g 2 − c = 2a ….(i) 2 f 2 − c = 2b ….(ii) On squaring (i) and (ii) and then subtracting (ii) from (i), we get g2 − f 2 = a2 − b 2. Hence the locus is x 2 − y 2 = a2 − b 2. www.sakshieducation.com  www.sakshieducation.com 20. (c) It is a fundamental concept. 21. (a) Equation of circle concentric to given circle is x 2 + y 2 − 6 x + 12 y + k = 0 ….(i) Radius of circle (i) = 2 (radius of given circle) ⇒ 9 + 36 − k = 2 9 + 36 − 15 ⇒ 45 − k = 60 ⇒ k = −15 Hence the required equation of circle is x 2 + y 2 − 6 x + 12 y − 15 = 0. 22. (a) According to the question, the required circle passes through (0,–1). Therefore, the radius is the distance between the points (0, –1) and (1, –2) i.e., 2. Hence the equation is ( x − 1) 2 + (y + 2) 2 = ( 2 ) 2 ⇒ x 2 + y 2 − 2x + 4 y + 3 = 0 23. (c) Conditions are g= f =r and g2 + f 2 − c = r ⇒ g = c. 24. (a) Here r = 10 (radius) Centre will be the point of intersection of the diameters, i.e. (8, –2). Hence required equation is (x − 8 ) 2 + (y + 2) 2 = 10 2 ⇒ x 2 + y 2 − 16 x + 4 y − 32 = 0. 25. (d) If c=0 ; circle passes through origin. 26. (b) We have the equation of circle x 2 + y 2 + 2 gx + 2 fy + c = 0 Y (–g,0) X′ X Y′ But it passes through (0, 0) and (2, 1), then c=0 ….(i) 5 + 4g + 2f = 0 ….(ii) Also g2 + f 2 − c =| g | ⇒ f =0 {∵ c = 0} 5 From (ii), g=− 4 www.sakshieducation.com  www.sakshieducation.com 27. (c) Radius = 2, C (0, 0) Y x = –1 x=1 X' X –1 1 O Y' Equation of circle can be found from this. 1 28. (a) The circle is x 2 + y2 − x =0. 2 1  1 1 Centre (− g, − f ) =  , 0  and R= +0−0 =. 4  16 4 29. (a) Required equation is (x − a) 2 + (y − a) 2 = a 2 ⇒ x 2 + y 2 − 2 ax − 2ay + a 2 = 0. a b a2 + b 2 30. (a) Centre is  ,  and radius = 2 2 4 Y (a/2, b/2) b a X O 31. (c) Obviously Y from figure, (0,3) r 3 O X 4 Radius is r = 42 + 32 = 5. 32. (c) Solving y=0 and y + 3x = 6 , we get (2 3 , 0) , only option (c) satisfies the co-ordinate.  n   n  33. (a) Line l1 x + m 1 y + n1 = 0 cuts x and y-axes in A  − 1 , 0  , B  0, − 1   and line l2 x + m 2 y + n2 = 0 cuts  l1   m1   n   − n2  axes in C  − 2 , 0  , D  0, .   2 l   m2  So AC and BD are chords along x and y-axes intersecting at origin O. Since A, B, C, D are concyclic, so OA.OC = OB.OD www.sakshieducation.com  www.sakshieducation.com  n1   n2   n  n  Or −  −  = − 1  − 2   l  l   m  m   1  2   1  2  Or | l1 l 2 | = | m 1 m 2 | So l1 l 2 = m 1 m 2 is correct among the given choices, which is given in (a). 34. (a) verification 35. (b) It is clear from the figure that diameter is m +n. X Y (m+n) B A O 90o n m 36. (a) Centre (–2, 1), radius = 36 + 4 = 40 Hence equation of circle is x 2 + y 2 + 4 x − 2 y − 35 = 0. 37. (d) ( x + 4 )( x − 12 ) + (y − 3 )(y + 1) = 0 38. (c) Radius = Distance from origin = α2 + β2 ∴ (x − α ) 2 + (y − β ) 2 = α 2 + β 2 ⇒ x 2 + y 2 − 2αx − 2 βy = 0. 39. (d) Let the centre be (h, k), then radius = h Also CC 1 = R1 + R 2 or (h − 3)2 + (k − 3)2 = h + 9 + 9 − 14 ⇒ (h − 3)2 + (k − 3)2 = h 2 + 4 + 4 h ⇒ k 2 − 10 h − 6 k + 14 = 0 or y 2 − 10 x − 6 y + 14 = 0. www.sakshieducation.com  www.sakshieducation.com 40. (c) Let AB be the chord of length 2 , O be centre of the circle and let OC be the perpendicular from O on AB. Then O 45° A C B AB = 2 2 1 AC = BC = = 2 2 1 In ∆ OBC , OB = BC cosec 45 o =. 2 =1 2 ∴ Area of the circle = π (OB )2 = π. 41. (a) It is obvious. 42. (c) The circle g, f, c passes through (2, 0) ∴ 4 + 4g + c = 0 ….(i) Intercept on x-axis is 2 (g 2 − c) = 5 ∴ 4 (g 2 + 4 g + 4 ) = 25 by (i) 9 1 Or (2 g + 9 )(2 g − 1) = 0 ⇒ g = − , 2 2 Since centre (− g, − f ) lies in 1st quadrant, we choose g=− 9 so that −g = 9 (positive). 2 2 ∴ c = 14 , (from (i)). 43. (c) Both the circles given in option (a) and (b) satisfy the given conditions. 44. (b) The centre of the circle which touches each axis in first quadrant at a distance 5, will be (5, 5) and radius will be 5. ∴ ( x − h) 2 + (y − k ) 2 = a 2 ⇒ ( x − 5 ) 2 + (y − 5 ) 2 = (5 ) 2 ⇒ x 2 + y 2 − 10 x − 10 y + 25 = 0. 45. (a) Centre is (1, − 1) (point of intersection of two given lines) and πr 2 = 154 ⇒ r = 7 ∴ Equation of required circle is ( x − 1)2 + (y + 1)2 = 49 ⇒ x 2 + y 2 − 2 x + 2 y − 47 = 0. www.sakshieducation.com  www.sakshieducation.com 46. (b) ∵ Radius of circle = perpendicular distance of tangent from the centre of circle 47. (d) Since circle touches the co-ordinate axes in III quadrant. r r (h1k) ∴ Radius = −h = −k. Hence h = k = −5 ∴ Equation of circle is (x + 5 )2 + (y + 5 )2 = 25. 48. (d) According to question two diameters of the circle are 2 x + 3y + 1 = 0 and 3x − y − 4 = 0 Solving, we get x = 1, y = −1 ∴ Centre of the circle is (1, – 1) Given 2πr = 10 π ⇒ r = 5 ∴ Required circle is (x − 1) 2 + (y + 1) 2 = 5 2 or x 2 + y 2 − 2 x + 2 y − 23 = 0. 49. (b) The equation of circle through points (0, 0), (1, 3) and (2, 4) is x 2 + y 2 − 10 x = 0 50. (d) Radius of given circle = g2 + f 2 − c g2 + f 2 = c (given), ∴ Radius = 0. 51. (d) Suppose the centre of circle be (h, k ). Since it touches the y - axis , ∴ radius of circle = h Now (h − 2)2 + k 2 = h 2 ⇒ h 2 + 4 − 4 h + k 2 = h 2 ⇒ k 2 = 4h − 4. Hence the locus of centre is y2 = 4 x − 4 , which is a parabola. 52. (d) Comparing the given equation with x 2 + y 2 + 2 gx + 2 fy + c = 0 , we get g=5 ∴ Length of intercept on x-axis = 2 g2 − c www.sakshieducation.com  www.sakshieducation.com = 2 (5) 2 − 9 = 8. 53. (b) x = 2 + 3 cos θ , y = 3 sin θ − 1 x 2 + y 2 = 4 + 9 cos 2 θ + 12 cos θ + 9 sin 2 θ + 1 − 6 sin θ = 14 + 12 cos θ − 6 sin θ = 4 (2 + 3 cos θ ) − 2(3 sin θ − 1) + 4 ⇒ x 2 + y 2 = 4 x − 2y + 4 ⇒ (x 2 − 4 x + 4 ) + (y 2 + 2 y + 1) = 9 ⇒ (x − 2)2 + (y + 1)2 = 9 , ∴ centre is (2,−1). 54. (b) The equation of circle passing through (0, 0), (2, 0) and (0, – 2) is x 2 + y 2 − 2 x + 2y = 0. If it passes through (k , − 2) , then k 2 + 4 − 2k − 4 = 0 ⇒ k = 0, 2 ∵ (0, − 2) is already a point on circle ∴ k = 2. 55. (a) x1 , x 2 are roots of x 2 + 2x + 3 = 0 ⇒ x1 + x 2 = −2 x1 + x 2 ∴ = −1 (x1+ x2)/2,(y1+y2)/2) 2 (x2, y2) Centre (x1, y1) y1 , y 2 are roots of y 2 + 4 y − 12 = 0 y1 + y 2 ⇒ y1 + y 2 = − 4 ⇒ = −2 2  x 1 + x 2 y1 + y 2  Centre of circle  ,  = (−1,−2).  2 2  56. (c) The other end is (t, 3 − t) So the equation of the variable circle is ( x − 1)( x − t ) + (y − 1)(y − 3 + t ) = 0 Or x 2 + y 2 − (1 + t )x − (4 − t)y + 3 = 0 ∴ The centre (α , β ) is given by www.sakshieducation.com  www.sakshieducation.com 1+t 4 −t α= ,β = 2 2 ⇒ 2α + 2 β = 5 Hence, the locus is 2 x + 2y = 5. 57. (b) Since the perpendicular drawn on chord from O(x , y ) bisects the chord. NM = a OM = y (ON )2 = (OM )2 + (ON )2 0(x,y) D x y x 2 = y 2 + a2 n a M x 2 − y 2 = a2 58. (b) According to the condition, (5 ) 2 + (3) 2 + 2(5 ) + k (3) + 17 = 7 ⇒ 61 + 3 k = 49 ⇒ k = −4. 59. (b) Here the equation of AB (chord of contact) is 0 + 0 − 3( x + 0 ) − 4 (y + 0 ) + 21 = 0 ⇒ 3 x + 4 y − 21 = 0 ….(i) A 2 O M C (0,0) (3,4) 2 B CM = perpendicular distance from (3, 4) to line (i) is 3 × 3 + 4 × 4 − 21 4 = 9 + 16 5 16 2 AM = AC 2 − CM 2 = 4 − = 21 25 5 4 ∴ AB = 2 AM = 21. 5 60. (a) Points where x +7 = 0 meets the circle x 2 + y 2 = 50 are (−7, 1) and (−7, − 1). Hence equations of tangents at these points are −7 x ± y = 50 or 7 x ± y + 50 = 0. 61. (c) According to the condition of tangency a cos α + b sin α − (a cos α + b sin α ) − r r= cos 2 α + sin 2 α ⇒ r =| −r | ⇒ r = r. www.sakshieducation.com  www.sakshieducation.com 62. (c) We know that the equation of normal to the circle x 2 + y 2 = a2 at the point (x 1 , y 1 ) is x y − =0. x 1 y1 x y Therefore , − =0 ⇒ x −y =0. 1/ 2 1/ 2 63. (a) Required equations are given by SS 1 = T 2 ⇒ (x 2 + y 2 − 2 x + 4 y )(1 + 4 ) = {y − 1(x ) + 2(y + 1)} 2 ⇒ 2 x 2 − 2 y 2 − 3 x + 4 y + 3 xy − 2 = 0 ⇒ (2 x − y + 1)( x + 2 y − 2) = 0. 64. (c) y = mx + c is a tangent, if c = ±a 1 + m 2 , where m = tan 45 o = 1 ∴ The equation is y = x ±6 2. 65. (c) Length of tangent is given by L T = S 1 = 49 = 7. 66. (b) If the line lx + my − 1 = 0 touches the circle x 2 + y 2 = a2 , then applying the condition of l.0 + m.0 − 1 tangency, we have ± =a l2 + m 2 1 On squaring and simplifying, we get the required locus x2 + y2 =. Hence it is a circle. a2 67. (a) The abscissa of point is found by substituting the ordinates and solving for abscissa. ⇒ x 2 − 8 x + 15 = 0 8 ± 64 − 60 8±2 ⇒x = = =5 or 3 2 2 i.e., points are (5, − 1) and (3, –1). x −5 y +1 Normal is given by, = ⇒ 2x + y − 9 = 0 5 − 4 −1 −1 x −3 y +1 and = ⇒ 2x − y − 7 = 0. 3 − 4 −1 −1 68. (c) k = 3, as perpendicular from centre on line = radius. f 2 + g2 − 6 4 69. (c) According to the condition, = ⇒ f 2 + g2 + 4 f + 4g + 2 = 0. f + g 2 + 3 f + 3g 2 1 70. (b) Normal passes through centre, therefore b = ma + c. 71. (a) From formula of tangent at a point,  ab 2   a 2b  2 2 x  2  + y  = a b ⇒ x + y =1.   a2 + b 2  a + b2 a +b 2 2    a b www.sakshieducation.com  www.sakshieducation.com 72. (a) The equation of tangents will be c(x 2 + y 2 + 2 gx + 2 fy + c) = (gx + fy + c) 2 These tangents are perpendicular, hence the coefficients of x2+ coefficients of y2 = 0 ⇒ c − g 2 + c − f 2 = 0 ⇒ f 2 + g 2 = 2c. 73. (d) Its centre is of type (c, c) and radius is 4 c + 3 c − 12 = c2 ⇒ c = 6. 5 74. (a) Tangent is x − 2y − 5 = 0 and points of intersection with circle x 2 + y 2 − 8 x + 6 y + 20 = 0 are given by 4 y 2 + 25 + 20 y + y 2 − 16 y − 40 + 6 y − 20 = 0 ⇒ 5 y 2 + 10 y + 5 = 0 75. (b) Length of tangent is S1. r2 Equation of circle is x2 + y2 − =0 a r2 Hence S1 = α 2 + β 2 −. a 76. (c) Centres of circles are C1 (2, 3) and C 2 (−3, − 9) and their radii are r1 = 5 and r2 = 8. Obviously r1 + r2 = C1 C 2 i.e., circles touch each other externally. Hence there are three common tangents. 77. (a) T ≡ x + 3y − 4 = 0 (1,√3) O (4,0) 1 Hence the required area = ×4× 3 = 2 3. 2 78. (d) Suppose that the point be (h, k). Tangent at (h, k) is hx + ky = a 2 ≡ x − y = − 2 a h k a2 a a or = = or h=− ,k = 1 − 1 − 2a 2 2  a a  Therefore, point of contact is  − , .  2 2  www.sakshieducation.com  www.sakshieducation.com 79. (a) concept 3 y + 10 80. (a) Substituting x = in equation of circle, we get a quadratic in y. Solving, we get two 4 values of y as 2 and – 6 from which we get value of x. θ CT 1 a 81. (c) tan = = 2 PT1 α + β 2 − a2 2 T2 θ C θ P(α,β) T1 82. (d) Equation of a tangent at (a cos α , a sin α ) to the circle x 2 + y 2 = a2 is ax cos α + ay sin α = a 2. a cos α Hence its gradient is − = − cot α. a sin α 83. (c) Substituting equation of line y = mx + c in circle x 2 + y2 = r2 84. (b) Area of quadrilateral = 2 [area of ∆OAC ] 1 = 2. OA. AC = S 1. g 2 + f 2 − c 2 A O C B  1   3 85. (b) Points of intersection with co-ordinate axes are  − , 0  , (0, 1) and (−3, 0),  0, .  λ   2  3 7 x 5y 3 Equation of circle through (0, 1), (–3, 0) and  0,  is x2 + y2 + − + =0.  2 2 2 2 86. (d) Tangent is of form 4 x + 3y + c = 0. From condition of tangency to the circle, we get c = −25. Hence equation is 4 x + 3 y − 25 = 0. 87. (c) Accordingly, α 2 + β 2 − 4α − 5 = α 2 + β 2 + 6α − 2 β + 6 ⇒ 10 α − 2 β + 11 = 0. 88. (a) Any tangent to x 2 + y2 = b2 is y = mx − b 1 + m 2. It touches ( x − a)2 + y 2 = b 2 , ma − b 1 + m 2 if =b or ma = 2b 1 + m 2 m +1 2 www.sakshieducation.com  www.sakshieducation.com 2b or m 2a 2 = 4 b 2 + 4 b 2m 2 , ∴ m = ±. a − 4b 2 2 89. (b) Let point of contact be P( x 1 , y 1 ). P(x1,y1) O(–1,1) 90. (c) Let P( x 1 , y 1 ) be a point on x2 + y2 = 4 91. (b) Put point (1, 2) in each option, only equation h + 2k = 5 satisfies. Hence option (b) is correct. 92. (c) Length of intercepted part is diameter i.e., 2r. 93. (a) The equation of tangent at point (−2, − 3) to the circle x 2 + y 2 + 2x + 4 y + 3 = 0 is, −2 x − 3 y + 1( x − 2) + 2(y − 3) + 3 = 0 ⇒ −2 x − 3 y + x − 2 + 2 y − 6 + 3 = 0 ⇒ −x − y − 5 = 0 ⇒ x + y + 5 = 0 or y = −x − 5 ; so, m = −1 −1 Hence, gradient of normal = =1. −1  1  a 94. (b) Obviously the slope of the tangent will be −  i.e., −. b/a b Y (a,b) X O a Hence the equation of the tangent is y=− x b 95. (b) y = − β x+β touches the circle, α  β 2  ∴ β 2 = a 2  1 + ⇒ 1 + 1 = 1  α 2  α 2 β 2 a2 2 1 1 1 ∴ Locus of  ,  is x 2 + y 2 =  .  α β  a www.sakshieducation.com  www.sakshieducation.com 96. (a) Both the sentences are true and R is the correct explanation of A, because for tangents dy which are parallel to x- axis, =0. dx 97. (a) Given tangents are 5 x − 12 y + 10 = 0, 5 x − 12 y − 16 = 0 c1 − c 2 26 Radius = = =1. 2 a +b 2 2 2.13 98. (c) Length of tangent = 3 2 + (−4 )2 − 4 (3) − 6 (−4 ) + 3 = 40 ∴ Square of length of tangent = 40. 99. (d) Let the point be (x1 , y1 ) x 12 + y12 + 4 x 1 + 3 2 According to question, = x12 + y12 − 6 x1 + 5 3 x 12 + y12 + 4 x 1 + 3 4 Squaring both sides, = x 12 + y12 − 6 x 1 + 5 9 ⇒ 9 x 1 + 9 y12 + 36 x 1 + 27 = 4 x 12 + 4 y12 − 24 x 1 + 20 ⇒ 5 x 12 + 5 y12 + 60 x 1 + 7 = 0 Hence, locus is 5 x 2 + 5 y 2 + 60 x + 7 = 0. 100. (a) Chord of contact from origin ≡ gx + fy + c = 0 and from (g, f ) ≡ gx + fy + g(x + g) + f (y + f ) + c = 0 Or 2 gx + 2 fy + g 2 + f 2 + c = 0 g2 + f 2 + c −c 2 g2 + f 2 − c ∴ Distance = =. g2 + f 2 2 g2 + f 2 101. (d) the point (1, 1) lies outside the circle, therefore no such chord exist. 102. (b) Since locus of middle point of all chords is the diameter, perpendicular to the chord. 103. (a) We know that the equation of common chord is S1 − S 2 = 0 , where S1 and S2 are the equations of given circles, therefore (x − a) 2 + (y − b ) 2 + c 2 − (x − b ) 2 − (y − a) 2 − c 2 = 0 ⇒ 2bx − 2 ax + 2 ay − 2by = 0 ⇒ 2(b − a)x − 2(b − a)y = 0 ⇒ x − y = 0. 104. (b) Let pole be (x1 , y1 ) then polar l m will be xx 1 + yy1 = 1 comparing with lx + my + n = 0 ⇒ x 1 = − , y1 = −. n n www.sakshieducation.com  www.sakshieducation.com 105. (b) Length of chord = 2{(radius) 2 − (length of ⊥from centre to chord )2 }1 / 2 1/2     2  2  −1   = 2 r −     (1 / a 2 ) + (1 / b 2 )       r 2 (a 2 + b 2 ) − a 2 b 2 =2. a2 + b 2 106. (b) Suppose point be (h, k). Equation of common chord of contact is hx + ky − a 2 = 0 ≡ lx + my + n = 0 h k − a2 − a2l − a2m or = = or h= , k =. l m n n n 107. (b) Equation of common chord PQ is 2x + 1 = 0 − 2 +1 1 C1 M = perpendicular distance of common chord from centre C1 = = − 2 2 2 P (–1,–3/2) M C2 C1 Q  3 3 Here; C1  − 1,−  , r1 = = C1 P  2 2 9 1 PQ = 2PM = 2 C1 P 2 − C1 M 2 = 2 − =2 2. 4 4 108. (d) Equation of common chord is S1 − S 2 = 0 ⇒ 2 x − 2y = 0 i.e., x−y =0 1 ∵ Length of perpencicular drawn from C1 to x−y =0 is 2 19 1 ∴ Length of common chord =2 − =6. 2 2 x ' y' a2 109. (a) Polar of the circle is xx '+ yy ' = a 2 , but it is given by Ax + By + C = 0 , then = = A B −C  a2 A a2B  Hence pole is    −C , −C .   c 110. (b) We know CD = ….(i) 1+m2 But according to figure, www.sakshieducation.com  www.sakshieducation.com a 2 − b 2 = CD 2 C (0,0) a b c2 From (i) and (ii), D we get a2 − b 2 = A y=mx+c (1 + m 2 ) ⇒ (a 2 − b 2 )(1 + m 2 ) = c 2. 111. (c) The centre of the given circle is (1, 3) and radius is 2. So, AB is a diameter of the given circle has its midpoint as (1, 3). The radius of the required circle is 3. 112. (d) Given, circle is x 2 + y 2 − 2x = 0......(i) and line is y=x.....(ii) Puting y=x in (i), B We get 2 x − 2 x = 0 ⇒ x = 0, 1 2 y=x From (i), y =0 ,1 H A Let A = (0, 0 ) , B = (1, 1) Equation of required circle is (x − 0) (x − 1) + (y − 0) (y − 1) = 0 or x2 + y2 − x − y = 0. 113. (d) Let chord AB is y = mx.....(i) Equation of CM, x + my = λ It is passing through (a, 0) ∴ x + my = a.....(ii) Y M B x2 + y2 – 2ax = 0 (0, 0) C (a, 0) A = a ⇒ x 2 + y 2 = ax y From (i) and (ii), x + y. x 114. (c) The given circle is x 2 + y 2 − 2x = 0. Let (x 1 , y 1 ) be the middle point of any chord of this circle, than its equation is S1 = T. or x 12 + y 12 − 2 x 1 = xx 1 + yy 1 − ( x + x 1 ) If it passes through (0, 0), then x 12 + y 12 − 2 x 1 = − x 1 ⇒ x 12 + y 12 − x 1 = 0 www.sakshieducation.com  www.sakshieducation.com Hence the required locus of the given point (x 1 , y 1 ) is x2 + y2 − x = 0. 115. (c) Obviously BC = 2 C (0,0) 3 1 B A x – 2y – k = 0 0 − 2.0 − k Hence, ± = 2 ⇒ k = ± 10. 1 2 + (−2) 2 116. (c) T = S1 is the equation of desired chord, hence xx 1 + yy 1 − a 2 = x 12 + y 12 − a 2 ⇒ xx 1 + yy 1 = x 12 + y 12. 2 117. (d) Centre (0, 0), radius = 3a × = 2a. 3 Hence circle x 2 + y 2 = 4a2 as centroid divides median in the ratio of 2 :1.  1 118. (a) Let equation of circle be x 2 + y 2 + 2 gx + 2 fy + c = 0. If  m,  lies on this circle, then  m 1 1 m2 + 2 + 2 gm + 2 f +c =0 m m or m 4 + 2 gm 3 + 2 fm + cm 2 + 1 = 0 This is a fourth degree equation in m having m1 , m 2 , m 3 , m 4 as its roots. 1 Therefore, m 1 m 2 m 3 m 4 = product of roots = =1. 1 a 119. (c) Required area = (h 2 + k 2 − a 2 ) 3 / 2 h +k2 2 3 192 = (4 2 + 3 2 − 9 ) 3 / 2 =. 4 2 + 32 25 120. (b) Let the equation of line passing through origin be y = mx. Therefore x 2 + y 2 − x + 3 y = 0 ⇒ x 2 + m 2 x 2 − x + 3 mx = 0 ⇒ x [ x (1 + m 2 ) − (1 − 3 m )] = 0 121. (b) Let A = (a, 0) , B = (−a, 0) , P = (α , β ) PA 2 ∴ = k2 ⇒ (α − a) 2 + β 2 = k 2 [(α + a) 2 + β 2 ] PB 2 ∴ Locus is ( x 2 + y 2 )(1 − k 2 ) − 2 a(1 + k 2 )x + (1 − k 2 )a 2 = 0