1. The centre (C) and radius (r) of the circle 4x2 + 4y2 - 10x + 5y + 5 = 0 are
Ans: (2)
Explanation:
2. A circle has its centre at (a, b). If the radius is a + b, its equation is
1) x2 + y2 - ax - by - ab = 0 2) x2 + y2 + ax + by + ab = 0
3) x2 + y2 - 2ax - 2by - 2ab = 0 4) x2 + y2 + 2ax + 2by + 2ab = 0
Ans: (3)
Explanation: (x - a)2 + (y - b)2 = (a + b)2
x2 + y2 - 2ax - 2by + a2 + b2 = a2 + b2 + 2ab
... x2 + y2 - 2ax - 2by - 2ab = 0
3. Area of the triangle formed by the center of the circles
x2 + y2 = 1, x2 + y2 + 6x - 2y = 1, x2 + y2- 12x + 4y = 1 is
1) 0 2) 1 3) 3 4) 6
Ans: (1)
Explanation: A = (0, 0), B = (-3, 1), C = (6, -2)
4. A circle with centre at (2, 3) passes through (2, 4). Its equation is
1) x2 + y2 - 4x - 6y + 12 = 0 2) x2 + y2 - 6x - 4y + 12 = 0
3) x2 + y2 + 4x + 6y - 12 = 0 4) x2 + y2 + 6x + 4y - 12 = 0
Ans: (1)
Explanation: C = (2, 3)
Circle equation: (x - 2)2 + (y - 3)2 = 12
x2 + y2 - 4x - 6y + 4 + 9 = 1
x2 + y2 - 4x - 6y + 12 = 0
5. A circle with its centre at (4, 5) passes through the centre of the circle x2 + y2 + 4x - 6y - 12 = 0. Its equation is
1) x2 + y2 + 8x + 10y + 1 = 0 2) x2 + y2 + 8x - 10y - 1 = 0
3) x2 + y2 - 8x + 10y - 1 = 0 4) x2 + y2 - 8x - 10y + 1 = 0
Ans: (4)
Explanation: Centre of the given circle = (-2, 3)
Centre of the required circle = (4, 5)
... Equation of the required circle:
(x - 4)2 + (y - 5)2 = 
x2 + y2 - 8x - 10y + 41 - 40 = 0 x2 + y2 - 8x - 10y + 1 = 0
6. A circle centered at (2, 3) passes through the intersection of the lines 4x + y = 27 and 3x - 2y - 1 = 0. The radius of the circle is
1) 2 2) 3 3) 4 4) 5
Ans: (4)
Explanation: Point of intersection of 4x + y = 27 and 3x - 2y = 1 is (5, 7)
Centre of the circle = (2, 3)
7. A diameter of the circle x2 + y2 - 6x + 2y - 8 = 0 passes through origin. Its equation is
1) x + 2y = 0 2) x - 2y = 0 3) x + 3y = 0 4) x - 3y = 0
Ans: (3)
Explanation: C = (3, -1)
O = (0, 0)
8. A circle is concentric with the circle x2 + y2 - 4x - 2y - 4 = 0. If it passes through the center of the circle x2 + y2 + 2x + 4y = 0, its equation is
1) x2 + y2 - 4x - 2y + 13 = 0 2) x2 + y2 - 4x - 2y - 13 = 0
3) x2 + y2 - 4x - 2y + 12 = 0 4) x2 + y2 - 4x - 2y - 12 = 0
Ans: (2)
Explanation: Required circle: x2 + y2 - 4x - 2y + k = 0
Centre of given circle = (-1, -2)
... 1 + 4 + 4 + 4 = -k
... k = -13
... Equation required: x2 + y2 - 4x - 2y - 13 = 0
9. The equation of the circle concentric with the circle x2 + y2 - 6x + 12y + 15 = 0 and of double its radius is
1) x2 + y2 - 6x + 12y - 75 = 0 2) x2 + y2 - 6x + 12y + 75 = 0
3) x2 + y2 - 6x + 12y + 65 = 0 4) x2 + y2 - 6x + 12y - 65 = 0
Ans: (1)
Explanation:
10. A circle passes through (1, 1), (2, -1), (3, -2). Its radius is
Ans: (3)
Explanation:
... Equation of circle: x2 + y2 - 13x - 5y + 16 = 0
11. The equation of the circle passing through the points (1, 2), (3, 6), (5, -6) is
1) x2 + y2 + 22x - 4y + 15 = 0 2) x2 + y2 - 22x + 4y + 15 = 0
3) x2 + y2 + 22x + 4y + 25 = 0 4) x2 + y2 - 22x - 4y + 25 = 0
Ans: (4)
Explanation: 2g + 4f + c = -5; 6g + 12f + c = -45; 10g - 12f + c = -61
Solving g = -11, f = -2, c = 25
... Circle equation: x2 + y2 - 22x - 4y + 25 = 0
12. If a circle passes through (5, 7), (6, 6), (2, -2), its centre and radius are
1) (2, 3), 5 2) (3, 2), 5 3) (2, 5), 3 4) (3, 5), 2
Ans: (1)
Explanation: 10g + 14f + c = -74
12g + 12f + c = -72
4g - 4f + c = -8 g = -2, f = -3, c = -12
... C = (-g, -f) = (2, 3)
13. A circle has its centre on the line 4x + y - 16 = 0. If it passes through the points (6, 5) and (4, 1), its equation is
1) x2 + y2 - 6x + 8y + 25 = 0 2) x2 + y2 + 6x - 8y - 25 = 0
3) x2 + y2 - 6x - 8y + 15 = 0 4) x2 + y2 + 6x + 8y - 15 = 0
Ans: (3)
Explanation: 4(-g) + (-f) - 16 = 0 ... 
(6, 5) 
(4, 1) 
Solving, g = -3, f = -4, c = 15
... Equation of circle: x2 + y2 - 6x - 8y + 15 = 0
14. A circle passing through (4, 2) and (-6, -2) has its centre on Y-axis. Its equation is
1) x2 + y2 + 5y + 30 = 0 2) x2 + y2 + 5y - 30 = 0
3) x2 + y2 - 5y + 30 = 0 4) x2 + y2 - 5y - 30 = 0
Ans: (2)
Explanation: 8g + 4f + c = -20
-12g - 4f + c = -40
(-g, -f) lies on Y-axis g = 0
on solving, we get 2f = 5; c = -30
... Circle required: x2 + y2 + 5y - 30 = 0
15. A circle passes through the points (h, k), (7, 0), (5, 2) and (1, -6) then h + k + 5 =
1) -1 2) -4 3) -5 4) 0
Ans: (4)
Explanation: 
g = -3, f = 2, c = -7
also (h, k) = (-1, -4) h + k + 5 = 0
16. The equation of the circle with radius 1 and passing through (1, 1) , (2, 2) is
1) x2 + y2 - 4x - 2y + 4 = 0 2) x2 + y2 + 4x + 2y + 4 = 0
3) x2 + y2 - 2x - 4y - 4 = 0 4) x2 + y2 + 2x + 4y - 4 = 0
Ans: (1)
Explanation: 
g2 + f2 = 1 + c
2g + 2f + c = -2
4g + 4f + c = -8
solving, we get g = -2, f = -1, c = 4
... Circle equation: x2 + y2 - 4x - 2y + 4 = 0
17. A circle of 5 radius has its centre on X-axis. If it passes through (2, 3), its equation is
1) x2 + y2 - 12x - 11 = 0 2) x2 + y2 + 12x + 11 = 0
3) x2 + y2 - 4x + 21 = 0 4) x2 + y2 + 4x - 21 = 0
Ans: (4)
Explanation: 
5 g2 + f2 - c = 25
(-g, -f) on X-axis f = 0
(2, 3) 4g + 6f + c = -13
solving, we get g = 2, c = -21
... Equation of circle = x2 + y2+ 4x - 21 = 0
18. The equation to the locus of the point of intersection of the lines x sin α - y cos α = b and x cos α + y sin α = a is
1) x2 + y2 = a2 2) x2 + y2 = b2 3) x2 + y2 = a2 + b2 4) xy = a2b2
Ans: (3)
Explanation: Squaring and adding, x2 + y2 = a2 + b2
19. The equation to the locus of the foot of the perpendicular from origin to the line which always passes through a fixed point (h, k) is
1) x2 + y2 + hx + ky = 0 2) x2 + y2 + hx - ky = 0
3) x2 + y2 - hx + ky = 0 4) x2 + y2 - hx - ky = 0
Ans: (4)
Explanation:
Ans: (2)
Explanation:
21. A circle passing through origin makes positive intercepts a and b on the coordinate axes. Its equation is
1) x2 + y2 + ax + by = 0 2) x2 + y2 + ax - by = 0
3) x2 + y2 - ax + by = 0 4) x2 + y2 - ax - by = 0
Ans: (4)
Explanation: When a circle makes positive intercepts a and b on the coordinate axes, we have the end points of the diameter as (a, 0) and (0, b).
... The equation of the circle: (x - a)(x) + (y)(y - b) = 0
x2 + y2 - ax - by = 0
22. A circle circumscribes a square ABCD of side 'a'. If AB and AD represent the coordinate axes, then the equation of the circle is
1) x2 + y2 + ax + ay = 0 2) x2 + y2 - ax - ay = 0
3) x2 + y2 + ax - ay = 0 4) x2 + y2 - ax + ay = 0
Ans: (2)
Explanation: When a circle circumscribes a square ABCD of side 'a' and AB and AD represent the coordinate axes, we have A = (0, 0), B = (a, 0), C = (a, a), D = (0, a).
... The equation of the circle with AC as diameter is
(x - 0)(x - a) + (y - 0)(y - a) = 0
x2 - ax + y2 - ay = 0 x2 + y2 - ax - ay = 0
23. A = (a, b), B = (c, d), P = (x, y) are three points. If 
is a right angle, the equation to the locus of P is
1) x2 + y2 - (a + c)x - (b + d)y + (ac + bd) = 0
2) x2 + y2 + (a + c)x + (b + d)y - (ac + bd) = 0
3) x2 + y2 - (a + b)x - (c + d)y + (ab + cd) = 0
4) x2 + y2 + (a + b)x + (c + d)y - (ab + cd) = 0
Ans: (1)
Explanation: (Slope of AP)(Slope of PB) = -1
(y - b)(y - d) + (x - a)(x - c) = 0 x2 + y2 - (a + c)x - (b + d)y + ac + bd = 0
24. (3, 4) is one end of diameter of a circle x2 + y2 - 4x - 6y + 11 = 0. The other end is
1) (1, 2) 2) (2, 3) 3) (3, 1) 4) (3, 2)
Ans: (1)
Explanation: C = (2, 3) is the centre.
A = (3, 4) is one end of the diameter.
B = (h, k) is the other end
(h, k) = (1, 2)
1) ab(x + y) = a2 + b2 2) (a2 + b2)(x + y) = a2b2
3) (bx + ay) = ab 4) (ax + by) = ab
Ans: (3)
Explanation:
26. Equation of normal to 2x2 + 2y2 + 3x - 4y + 1 = 0 at (-1, 2) is
1) x + y + 2 = 0 2) 4x + y + 2 = 0 3) x + 4y + 2 = 0 4) 4x + 2y + 3 = 0
Ans: (2)
Explanation:
27. The equation of normal to the circle x2 + y2 - 8x - 2y + 12 = 0 at the points whose ordinate -1 is
1) 2x - y -7 = 0 2) 2x - y + 7 = 0 3) 2x + y + 9 = 0 4) 2x - y - 9 = 0
Ans: (1)
Explanation: P = (h, -1) h2 + 1 - 8h + 2 + 12 = 0
h2 - 8h + 15 = 0
h = 3, 5
... P(3, -1) or (5, -1)
28. A line y = mx + c is a normal to the circle with centre (a, b) and radius r. Then
1) am = b + c 2) am = b - c 3) bm = a + c 4) bm = a - c
Ans: (2)
Explanation: Normal always passes through centre of a circle.
... y = mx + c passes through (a, b)
b = m(a) + c (or) am = b - c.
29. The normal at (2, 3) to the circle x2 + y2 + 4x + 6y - 39 = 0 meets the circle also at
1) (-2, -3) 2) (-6, 9) 3) (6, -9) 4) (-6, -9)
Ans: (4)
Explanation: Normal passes through centre (-2, -3) of the circle.
Therefore, the normal becomes the diameter for which (2, 3) is one end.
If (h, k) is the other end, we have
... h = -6, k = -9
... Normal at (2, 3) meets the circle at (-6, -9).
30. A line parallel to x + 2y - 3 = 0 is a normal to the circle x2 + y2 - 2x = 0. Its equation is
1) x + 2y - 1 = 0 2) x + 2y + 1 = 0 3) x + 2y + 2 = 0 4) x + 2y - 2 = 0
Ans: (1)
Explanation: x + 2y - k = 0 is parallel to x + 2y - 3 = 0
Passes through centre (1, 0) of the circle.
1 -k = 0
... k = 1
... The required equation is x + 2y - 1 = 0.
31. y = mx + c touches x2 + y2 = r2 at
Ans: (2)
Explanation: xx1 + yy1 = r2 tangent at (x1, y1)
mx - y = -c given tangent
32. y = x + 
touches x2 + y2 = a2 at
Ans: (3)
Explanation:
33. The area of the triangle formed by the tangent at (α, β) to the circle x2 + y2 = r2 with the coordinate axes is
Ans: (2)
Explanation:
34. 2b is the intercept made by the circle x2 + y2 = a2 with the line y = mx + c. Then c2 =
1) a2(1 + m2) 2) b2(1 + m2) 3) (a2 + b2)(1 + m2) 4) (a2 - b2)(1 + m2)
Ans: (4)
Explanation:
35. A line y = x + 2 is cut off by the circle x2 + y2 + 4x - 2y - 3 = 0. The middle point of the line is:
Ans: (2)
Explanation: Middle point of the line is the foot of the perpendicular from the centre of the circle.
36. The point of contact of the line 3x + 4y + 7 = 0 with the circle x2 + y2 - 4x - 6y - 12 = 0 is
1) (-1, -1) 2) (-1, 1) 3) (1, -1) 4) (1, 1)
Ans: (1)
Explanation:
37. A tangent to the circle x2 + y2 = 25 is inclined at an angle of 60° with the X - axis. Its one equation is:
Ans: (4)
Explanation:
38. A line perpendicular to 4x + 3y = 7 touches the circle x2 + y2 - 4x - 6y - 12 = 0. The equation of the line is
1) 3x - 4y - 19 = 0 2) 3x - 4y + 19 = 0 3) 3x + 4y - 31 = 0 4) 3x + 4y + 31 = 0
Ans: (1)
Explanation: line perpendicular to 4x + 3y = 7 is 3x - 4y = k
touches the circle radius = perpendicular distance from centre
39. Tangents drawn to x2 + y2 = 10 intersect at (4, -2). The angle between them is
1) 30° 2) 45° 3) 60° 4) 90°
Ans: (4)
Explanation:
40. α, β are the inclinations of tangents drawn from P to the circle x2 + y2 = a2. If (tan α)(tan β) = k, the equation to the locus of P is
1) k(x2 - a2) = y2 + a2 2) k(x2 - a2) = y2 - a2
3) x2 - a2 = k(y2 + a2) 4) x2 - a2 = k(y2 - a2)
Ans: (2)
Explanation:
