Q: Let the circle x^ 2 +y^ 2 =4 intersect x-axis at the points A ( a , 0), a >0 and B ( b , 0). Let P (2 , 2 ), 0< < 2 and Q (2 , 2 ) be two points such that ( - )= 2 . Then the point of intersection of

Let point of intersection R(h,k) m_ BR = m_ BP k h+2 = 2 2 + 2 k h+2 = 2 m_ AR = m_ AQ k h-2 = 2 2 - 2 = - 1 = - 2 2 - 2 = 4 ( 2 - 2 ) = 4 = 1 2 - 2 1 + 2 2 = 1 k h+2 + h-2 k 1 + ( k h+2 ) ( 2-h k ) = 1 k^2 + h^2 - 4 k(h+2) 4 h+2 = 1 h^2 + k^2 - 4 4k = 1 x^2 + y^2 - 4y - 4 = 0

Q: Let y=x be the equation of a chord of the circle C _ 1 (in the closed half-plane x 0) of diameter 10 passing through the origin. Let C _ 2 be another circle described on the given chord as its diamete

Circle C_1 has radius 5 and chord on line y = x through origin. Setting center at (5, 0) (in region x 0), the chord endpoints are (0, 0) and (5, 5) by solving (t-5)^2 + t^2 = 25. Circle C_2 has this chord as diameter, so center is at ( 5 2 , 5 2 ) with radius 5 2 2 . The chord of C_2 through (2, 3) that is farthest from center is perpendicular to the radial direction from center to (2, 3). The radial direction is (2 - 5 2 , 3 - 5 2 ) = (- 1 2 , 1 2 ) with normal direction (1, -1). The chord equa

Question 1

Consider the circle C: x^2+y^2-6x-8y-11=0. Let a variable chord AB of the circle C subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord

Accepted Answer

Let the foot of the perpendicular from the origin to the chord AB be P(h, k). The slope of OP is k h . Since the chord AB is perpendicular to OP, its equation is given by: y - k = - h k (x - h) hx + ky = h^2 + k^2 This can be rewritten as hx + ky h^2 + k^2 = 1. The chord AB subtends a right angle at the origin. We homogenize the equation of the circle x^2 + y^2 - 6x - 8y - 11 = 0 with the equation of the chord to find the joint equation of the lines OA and OB: x^2 + y^2 - (6x + 8y) ( hx + ky h^2

Question 2

Let C be a circle having centre in the first quadrant and touching the x-axis at a distance of 3 units from the origin. If the circle C has an intercept of length 6 3 on y-axis, then the length of the

Accepted Answer

Let the centre of the circle be (3, r) since it touches the x-axis at (3, 0) and lies in the first quadrant. The radius of the circle is r. The length of the intercept on the y-axis is given by 2 r^2 - d^2 , where d is the distance of the centre from the y-axis. Here, d = 3. Given that the intercept on the y-axis is 6 3 : 2 r^2 - 3^2 = 6 3 r^2 - 9 = 3 3 r^2 - 9 = 27 r^2 = 36 r = 6 The centre of the circle is (3, 6) and its radius is 6. The given line is x - y - 3 = 0. The perpendicular distance

Question 3

Let the line x - y = 4 intersect the circle C: (x-4)^2 + (y+3)^2 = 9 at the points Q and R. If P( , ) is a point on C such that PQ = PR, then (6 + 8 )^2 is equal to __________.

Accepted Answer

The given circle C: (x-4)^2 + (y+3)^2 = 9 has its center at C_0(4, -3) and radius r = 3. The line QR has the equation x - y = 4, which has a slope of 1. Since PQ = PR, the point P( , ) must lie on the perpendicular bisector of the chord QR. The perpendicular bisector of any chord of a circle passes through its center. Therefore, the perpendicular bisector passes through C_0(4, -3) and has a slope of -1 (since it is perpendicular to QR). The equation of the perpendicular bisector is: y - (-3) = -

Question 4

Let the centre of the circle x^2 + y^2 + 2gx + 2fy + 25 = 0 be in the first quadrant and lie on the line 2x - y = 4. Let the area of an equilateral triangle inscribed in the circle be 27 3 . Then the

Accepted Answer

The equation of the circle is x^2 + y^2 + 2gx + 2fy + 25 = 0. The centre of the circle is (-g, -f). Since it lies in the first quadrant, -g > 0 and -f > 0, which implies g The centre lies on the line 2x - y = 4, so: -2g - (-f) = 4 f = 2g + 4 Let the radius of the circle be R. The side length a of an equilateral triangle inscribed in the circle is a = 3 R. The area of the equilateral triangle is given as 27 3 : 3 4 a^2 = 27 3 3 4 (3R^2) = 27 3 R^2 = 36 The radius of the circle is also given by R^

Question 5

Let P be a moving point on the circle x^2 + y^2 - 6x - 8y + 21 = 0. Then, the maximum distance of P from the vertex of the parabola x^2 + 6x + y + 13 = 0 is equal to:

Accepted Answer

The equation of the given circle is x^2 + y^2 - 6x - 8y + 21 = 0. The center of the circle is C(3, 4) and its radius is r = 3^2 + 4^2 - 21 = 4 = 2. The equation of the given parabola is x^2 + 6x + y + 13 = 0. Rewriting the equation by completing the square, we get (x + 3)^2 = -(y + 4). The vertex of the parabola is V(-3, -4). The maximum distance of a moving point P on the circle from the vertex V is given by CV + r. The distance between the center C(3, 4) and the vertex V(-3, -4) is CV = (3 - (

Question 6

Suppose that two chords, drawn from the point (1, 2) on the circle x^2 + y^2 + x - 3y = 0 are bisected by the y-axis. If the other ends of these chords are R and S, and the mid point of the line segme

Accepted Answer

Let the other end of the chord be (x_1, y_1). Since the chord is bisected by the y-axis, the midpoint of the chord lies on the y-axis. The midpoint of the chord joining (1, 2) and (x_1, y_1) is ( x_1 + 1 2 , y_1 + 2 2 ). Since it lies on the y-axis, its x-coordinate must be zero: x_1 + 1 2 = 0 x_1 = -1 Since the point (x_1, y_1) lies on the given circle x^2 + y^2 + x - 3y = 0, we substitute x_1 = -1 into the equation: (-1)^2 + y_1^2 + (-1) - 3y_1 = 0 1 + y_1^2 - 1 - 3y_1 = 0 y_1^2 - 3y_1 = 0 y_1

Question 7

Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines x + (k-1)y + 3 = 0 and 2x + k^2 y - 4 = 0. If the line x - y + 2 = 0 intersects the

Accepted Answer

The given lines are x + (k-1)y + 3 = 0 and 2x + k^2 y - 4 = 0. Since they are mutually perpendicular, the product of their slopes is -1: ( -1 k-1 ) ( -2 k^2 ) = -1 2 k^2(k-1) = -1 k^3 - k^2 + 2 = 0 By trial, k = -1 is a root. Factoring gives (k+1)(k^2 - 2k + 2) = 0. Since k^2 - 2k + 2 = 0 has no real roots, k = -1. Substituting k = -1 into the equations of the lines, we get: x - 2y + 3 = 0 2x + y - 4 = 0 Solving these two equations, we get the point of intersection as x = 1 and y = 2. Thus, the

Question 8

Let a circle C have its centre in the first quadrant, intersect the coordinate axes at exactly three points and cut off equal intercepts from the coordinate axes. If the length of the chord of C on th

Accepted Answer

Let the equation of the circle be x^2 + y^2 + 2gx + 2fy + c = 0. Since the circle intersects the coordinate axes at exactly three points, it must pass through the origin and intersect each axis at one other distinct point. Thus, c = 0. The lengths of the intercepts on the x-axis and y-axis are 2 g^2 - c = 2|g| and 2 f^2 - c = 2|f|. Given that the intercepts are equal, we have |g| = |f|. The centre of the circle is (-g, -f). Since it lies in the first quadrant, -g > 0 and -f > 0. Therefore, g = f

Question 9

Let the circle x^ 2 +y^ 2 =4 intersect x-axis at the points A ( a , 0), a >0 and B ( b , 0). Let P (2 , 2 ), 0< < 2 and Q (2 , 2 ) be two points such that ( - )= 2 . Then the point of intersection of

Accepted Answer

Let point of intersection R(h,k) m_ BR = m_ BP k h+2 = 2 2 + 2 k h+2 = 2 m_ AR = m_ AQ k h-2 = 2 2 - 2 = - 1 = - 2 2 - 2 = 4 ( 2 - 2 ) = 4 = 1 2 - 2 1 + 2 2 = 1 k h+2 + h-2 k 1 + ( k h+2 ) ( 2-h k ) = 1 k^2 + h^2 - 4 k(h+2) 4 h+2 = 1 h^2 + k^2 - 4 4k = 1 x^2 + y^2 - 4y - 4 = 0

Question 10

Let y=x be the equation of a chord of the circle C _ 1 (in the closed half-plane x 0) of diameter 10 passing through the origin. Let C _ 2 be another circle described on the given chord as its diamete

Accepted Answer

Circle C_1 has radius 5 and chord on line y = x through origin. Setting center at (5, 0) (in region x 0), the chord endpoints are (0, 0) and (5, 5) by solving (t-5)^2 + t^2 = 25. Circle C_2 has this chord as diameter, so center is at ( 5 2 , 5 2 ) with radius 5 2 2 . The chord of C_2 through (2, 3) that is farthest from center is perpendicular to the radial direction from center to (2, 3). The radial direction is (2 - 5 2 , 3 - 5 2 ) = (- 1 2 , 1 2 ) with normal direction (1, -1). The chord equa

Question 11

Accepted Answer

Question 12

Let a circle of radius 4 pass through the origin O, the points A (- 3 a, 0) and B (0,- 2 b), where a and b are real parameters and a b 0. Then the locus of the centroid of O A B is a circle of radius

Accepted Answer

Circle of radius 4 passes through O and points A(- 3a , 0), B(0, - 2b ). Let center be C(h, k) on circle: h^2 + k^2 = 16. Since A, B lie on circle: 3a^2 + 2 3a h + h^2 + k^2 = 16 gives h = - 3a 2 And h^2 + 2b + 2 2b k + k^2 = 16 gives k = - 2b 2 Centroid G = ( - 3a 3 , - 2b 3 ). Setting x = - 3a 3 and y = - 2b 3 : a = 3x^2 and b = 9y^2 2 From h^2 + k^2 = 16: 3a 4 + 2b 4 = 16 → 3a + 2b = 64 Substituting: 9x^2 + 9y^2 = 64 → x^2 + y^2 = 64 9 Radius = 8 3

Question 13

If the points of intersection of the ellipses x^ 2 +2 y^ 2 -6 x-12 y+23=0 and 4 x^ 2 +2 y^ 2 -20 x-12 y+35=0 lie on a circle of radius r and centre (a, b), then the value of a b+18 r^ 2 is

Accepted Answer

The circle through intersection of the ellipses is S_1 + S_2 = 0. (1+4 )x^2 + (2+2 )y^2 - (6+20 )x - (12+12 )y + (23+35 ) = 0. For a circle: 1+4 = 2+2 = 1 2 . Circle: 3x^2 + 3y^2 - 16x - 18y + 81 2 = 0 x^2 + y^2 - 16 3 x - 6y + 27 2 = 0. Centre (a,b) = ( 8 3 , 3 ), r^2 = 64 9 + 9 - 27 2 = 47 18 . ab + 18r^2 = 8 3 3 + 18 47 18 = 8 + 47 = 55.

Question 14

Let the set of all values of r, for which the circles (x+1)^ 2 +(y+4)^ 2 =r^ 2 and x^ 2 +y^ 2 -4 x-2 y-4=0 intersect at two distinct points be the interval ( , ). Then is equal to

Accepted Answer

Circle 1: center C_1 = (-1, -4), radius r. Circle 2: (x-2)^2 + (y-1)^2 = 9, center C_2 = (2, 1), radius 3. Distance d = 9+25 = 34 . For two intersection points: |r-3| r + 3 > 34 r > 34 -3 and |r-3| r ( 34 -3,\ 34 +3), so = 34 -3, = 34 +3. = ( 34 )^2 - 9 = 34 - 9 = 25.

Question 15

If P is a point on the circle x^ 2 +y^ 2 =4, Q is a point on the straight line 5 x+y+2=0 and x-y+1=0 is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such points P is \_\_

Accepted Answer

Mid point of PQ lies on x - y + 1 = 0 2 + 2 - 2 - 5 - 2 2 + 1 = 0 2 + - 2 + 5 + 2 + 2 = 0 - + 3 + 2 = 0 ...(1) Slope of PQ is -1 2 + 5 + 2 2 - = -1 2 + 5 + 2 = -2 + + + 2 + 1 = 0 ...(2) eliminate from (1) and (2) + 5 = 1 , [0, 2 ] 5 2 2 2 = 2 ^2 2 2 = 0 = 1 or 2 = 5 = - 12 13 Sum of all possible values of abscissa of point P is = 2 1 + 2 ( -12 13 ) = 2 13 13 times sum of all possible values of abscissa of point P is 2.

Question 16

Let c and d be vectors such that | c + d |= 29 and c (2 i +3 j +4 k )=(2 i +3 j +4 k ) d . If _ 1 , _ 2 ( _ 1 > _ 2 ) are the possible values of ( c + d ) (-7 i +2 j +3 k ), then the equation K ^ 2 x^

Accepted Answer

From c a = a d where a = 2 i + 3 j + 4 k , we get ( c + d ) a = 0. So c + d = t a . Given | c + d | = 29 : |t| 29 = 29 , so t = 1. ( c + d ) (-7 i + 2 j + 3 k ) = (-14 + 6 + 12) = 4. _1 = 4, _2 = -4. For circle: Coeff of x^2 = Coeff of y^2: K^2 = 3K - 2 K = 1, 2. Coeff of xy = 0: K^2 - 5K + 4 = 0 K = 1, 4. Common value: K = 1.

Question 17

Let PQ and MN be two straight lines touching the circle x^ 2 +y^ 2 -4 x-6 y-3=0 at the points A and B respectively. Let O be the centre of the circle and AOB = / 3. Then the locus of the point of inte

Accepted Answer

Circle: (x-2)^2 + (y-3)^2 = 16, center O = (2, 3), radius r = 4. For tangents from external point P touching at A and B with AOB = /3: In quadrilateral OAPB: APB = - /3 = 2 /3, so OPA = /3. In right triangle OAP: ( /3) = 4 OP . OP = 8 3 = 8 3 3 . Locus: (x-2)^2 + (y-3)^2 = 64 3 . 3(x^2 + y^2) - 12x - 18y - 25 = 0.

Circle — Previous Year Questions & Solutions