Q1. If (1, 2) lies on the circle. x 2 + y 2 + 2gx + 2fy + c = 0. which is concentric with the circle x 2 + y 2 +4x + 2y 5 = 0 then c =
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1 Q1. If (1, 2) lies on the circle x 2 + y 2 + 2gx + 2fy + c = 0 which is concentric with the circle x 2 + y 2 +4x + 2y 5 = 0 then c = a) 11 b) -13 c) 24 d) 100
2 Solution: Any circle concentric with x 2 + y 2 + 4x + 2y 5 = 0 is x 2 + y 2 + 4x + 2y + c = 0. Substituting (1, 2), Ans: (b) c = 0 c= -13
3 Q2. The centre of a circle is (3, -1) and it makes an intercept of 6 units on the line 2x 5y + 18 = 0. The equation of the circle is a) x 2 + y 2-6x + 2y 28 = 0 b) x 2 + y 2 6x + 2y + 28 = 0 c) x 2 + y 2 + 4x 2y + 24 = 0 d) x 2 + y 2 + 2x 2y - 12=0
4 Solution: AB = 6, OC perpendicular distance from (3, -1) to2x 5y + 18 = = = A B 2 2 r = OB= OC + BC = 29+ 9= 38 (x 3) 2 + (y + 1) 2 = 38 x 2 + y 2 6x + 2y 28 = 0 C O (3, -1) Ans: (a)
5 Q3.The four distinct points, (0, 0), (2, 0), (0, -2) and (k, 2) are concyclic if k = a) 2 b) -2 c) 0 d) 1
6 Solution: (2, 0) and (0, -2) are ends of the diameter Eqn is x 2 + y 2 ax by = 0 x 2 + y 2 2x + 2y = 0 Substituting (k, -2), k k 4 = 0 k 2 2k = 0, k =2 (k 0) Ans: (a)
7 Q4. The length of the intercept made by the circle x 2 + y x 12y 13 = 0 on the y axis is a) 1 b) 2 c) 3 d) 14
8 Solution: Length = 2 2 f - c = =14 Ans: (d)
9 Q5. The length of the chord x = 3y + 13 of the circle x 2 + y 2 4x + 4y + 3 = 0 is a) 2 5 b) 5 2 c) 3 5 d) 10
10 Solution: OB= radius of the circle = = 5 OC = perpendicular distance from (2, -2) to x - 3y 13 = = = A C O (2,-2) 2 2 AB = 2BC = 2 OB - OC = = 10 = Ans: (d) B
11 Q6. The area of the circle whose equations are given by x = cos θ and y = sinθ is a) 4π b) 6π c) 9π d) 8π
12 Solution: Given x - 3 = 2cosθ θ and y 1 = 2 sinθ. (x 3) 2 + (y 1) 2 = 4 radius = 2 Area = 4π Ans: (a)
13 Q7. A circle having area 9π sq. units touches both the coordinate axes in the third quadrant, then its equation is a) x 2 + y 2 6x 6y 9 =0 b) x 2 + y 2 + 6x + 6y + 9 =0 c) x 2 + y 2 6x + 6y + 9 = d) x 2 + y 2 + 6x 6y + 9 =0
14 Solution: Area = 9π πr 2 = 9π r 2 = 9 r = 3 Since circle touches the coordinate axes in the third quadrant, its centre must be (-3, -3) Eqn is (x h) 2 + (y k) 2 = r 2 (x + 3) 2 + (y + 3) 2 = 9 x 2 + y 2 + 6x + 6y + 9 = 0 Ans: (b)
15 Q8. If x + y + k = 0 is a tangent to the circle x 2 + y 2 2x 4y + 3 = 0 then k = a) ± 20 b) -1,-5 c) ± 2 d) 4
16 Solution:- (1, 2) radius = = 2 x + y + k = 0 Perpendicular distance from (1, 2) to x + y + k = 0} = radius 1 2+ k 2 + = 2 K + 3 = ±2 k = -1, -5 Ans(b)
17 Q9. The radius of any circle touching the lines 3x - 4y + 5 = 0 and 6x 8y 9 =0 is a) 1 b) c) d) 19 20
18 Solution: 3x 4y + 5 = x 8y 9 = 0 2 Multiply 1 by 2 6x 8y + 10 = 0 Given lines are parallel lines 2r = distance between the lines c-d a + b r= = = r = Ans: (d)
19 Q10. The circles x 2 + y 2 10x + 16 = 0 and x 2 + y 2 = r 2 intersect each other in distinct points if a) r > 8 b) r < 2 c) 2 < r < 8 d) 2 r 8
20 Solution: C 1 = (5, 0), r 1 = 3, C 2 = (0, 0), r 2 = r Circles intersect if r 2 r 1 < c 1 c 2 < r 1 + r 2 r 3 < c 1 c 2 < 3 + r r 3 < 5 < 3 + r r 3 < 5 r < 8 r + 3 > 5 r > 2 2 < r < 8 Ans: (c)
21 Q11. If ax 2 + by 2 + (a + b 4) xy ax by 20 = 0 represents a circle, then its radius is a) 21 2 b) 42 2 c) 2 21 d) 22
22 Solution: Since the equation represents a circle. Coefficient of x 2 and y 2 are equal a = b Also xy coefficient = 0 a + b 4 = 0 2a 4 = 0 a = 2 b = 2 equation is 2x 2 + 2y 2 2x 2y 20 = 0 i.e. x 2 + y 2 x y 10 = 0 r = = Ans: (b)
23 Q12. If the centroid of an equilateral triangle is (1, 1) and one of the vertex is (-1, 2). Then the equation of the circumcircle is a) x 2 + y 2-2x - 2y - 3 = 0 b) x 2 + y 2 + 2x 2y - 3 = 0 c) x 2 + y 2 + 2x + 2y - 3 = 0 d) x 2 + y 2-2x + 3y - 3 = 0
24 Solution: Centroid = circumcentre centre = (1, 1) = 5 Radius = ( ) ( ) (x 1) 2 + (y 1) 2 = ( 5 ) 2 x 2 + y 2 2x 2y = 0 x 2 + y 2 2x 2y 3 = 0 Ans: (a)
25 Q13. The lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b =0 are concurrent if a) 2 a = 3abc b) a=0 c) a 2= ab d) 2 ) ) a =0
26 Solution: a b c b c a=0 c a b a (bc a 2 ) b (b 2 ca) + c (ab c 2 ) = 0 abc a 3 b 3 + abc + abc c 3 = 0 a 3 + b 3 + c 3 = 3abc a + b + c = 0 Σ a = 0 Ans: (b)
27 Q14. The area of the triangle whose sides are along the lines x = 0, y = 0 and 4x + 5y 20 = 0 is a) 20 sq. units b) 10 sq. units c) 1/10 sq. units d) 1/20 sq. unit
28 Solution: x = 0 B (0, 4) 4x + 5y = 20 O A (5, 0) y = 0 Are of OAB = 1 OA.OB = = Ans s(b)
29 Q15. The line x + y = 4 divides the line joining the points (-1, 1) and (5, 7) in the ratio a) 2 : 3 b) 1 : 2 c) 1 : 1 d) 4 : 3
30 Solution: Take A= (-1, 1) and B = (5, 7) Let x + y 4 =0 divide AB in the ratio k : 1 mx +nx my +ny P=, = 5k-1, 7k m+n m+n k+1 k+1 Substituting in x + y 4 = 0, we get 5k-1+ 7k+1-4=0 k+1 k+1 5k 1+7k+1-4k-4=0 8k = 4 1 k= 2 the ratio is k:1 = 1:2 Ans: (b)
31 Or m -ax+by +c = 1 1 n ax +by +c 2 2 m= = 4= 1 n m:n = 1:2 Ans: (b)
32 Q16. If p is the length of the perpendicular from the origin on the line whose intercepts on the axes are a and b, then 1 = p 2 a) a b 2 b) 2. a b 2 c) a 2 b 2 b 2 a d) 2
33 Solution: (0, b) (a, 0) Equation is x y + = 1 a b Length = c p= -1 a 2 +b 2 a +b p= ab p 2 = a b 1 = a +b = a+b a+b p a b b a 2 Ans: (a)
34 Q17. The reflection of the point (4, 4) with respect to the line x + y = 2 is a) (1, 1) b) (- 2, - 2) c) - 1, d) (-2, - 2)
35 Solution: (4, 4) (h, k) x + y 2 Mid point (4,4) & (h, k) = h+4 k+4, 2 2 Substituting in x + y 2 = 0, we get h+4+ k+4-2=0 2 2 Also k-4-1=-1 h-4 h = k.. (1) h k = 0 h + k = -4 (2) Solving (1) & (2), we get h=-2, k= -2 Ans. (d)
36 Q18. The point of intersection ti of pair of lines 2x 2 5xy 3y 2 + 6x +17y 20 = 0 is a) )(2, 1) b) )(1, 2) c) (1, -2) d) (-2, 1)
37 Solution:- Point of intersection hf -bg gh-af =, bh 2 bh 2 ab-h ab-h (3) 3-2 = 2 2, , = = (1, , =
38 Or Using partial derivative method, Difft. W.r.to x, keeping y constant 4x 5y +6 = 0 Difft. W. r. to y, keeping x constant -5x 6y + 17 = 0 Solving x = 1, and y = 2 Ans:- (b)
39 Q19. If the point of intersection of the lines kx+4y+2 = 0, x 3y+5 = 0 lie on 2x+7y 3 =0, then k= a) 2 b) 3 c) -2 d) -3
40 Solution :- Solving, x 3y + 5 = 0 and 2x + 7y 3 = 0 We get, Substituting x= -2 and y =1 in kx + 4y +2 =0, we get, y = 1, x = -2-2k = 0 Ans:- (b) k = 3
41 Q20. The equation of the line perpendicular to 5x 2y -7 = 0 and passing through the point of intersection of the lines 2x + 3y 1 = 0 and 3x + 4y 6 =0 is a) 2x + 5y + 17 = 0 b) 2x + 5y 17 = 0 c) 2x -5y + 17 = 0 d) 2x 5y 17 = 0
42 Solution:- Solving, 2x + 3y = 1 3x + 4y = x = 14,,y = -9
43 Line perpendicular p to 5x 2y 7 = 0 is 2x + 5y + k = 0 Substituting (14, -9 ) We get k = 0 k = 17 Equation is 2x+5y+17 = 0 Ans:- (a)
44 Q21. If (-4, 5) is one vertex and 7x y + 8 = 0 is one diagonal of a square then the equation of second diagonal is a) x + 3y 21 = 0 b) 2x - 3y - 7 = 0 c) x + 7y - 31 = 0 d) 2x + 3y - 21 = 0
45 Solution: The second diagonal passes through (-4, 5) and perpendicular to 7x y + 8 = 0 Equation is x + 7y +k =0 Substituting (-4, 5), k = 0 k = -31 Equation is x+ 7y - 31 = 0. Ans:- (c)
46 Q22. The distance between the pair of parallel lines 9x 2 6xy + y x 6y + 8 = 0 is a) 1 b) c) 4 d) 10 10
47 Solution:- distance = 2 2 g ac = aa+ b 9 10 = 2 1 = Ans:- (b)
48 Q23. If the sum of the slopes of the lines given by x 2 2k xy 7y 2 = 0 is four times their product then k = a) 1 b)-1 c) 2 d) -2
49 Solution:- m 1 + m 2 = 4. m 1 m 2 2h = 4. a - 2h = 4a b b Ans:- (c) -2(-k) = 4 k = 2
50 Q24. The area of the largest square inscribed in the circle x 2 + y 2 6x 8y = 0 is a) 100 sq units b) 25 sq units c) 10 sq units d) 50 sq units
51 Solution:- g = -3., f = -4, c = 0 5 a a Ans:- (d) Radius = 9+16 = 5 a 2 + a 2 = a 2 = 100 a 2 = 50 Area = 50 sq units
52 Q25. If 2y + x + 3 = 0 is a tangent to 5x 2 + 5y 2 = k, then the value of K is a) 4 b) 9 c) 16 d) 25
53 Solution :- 5x 2 + 5y 2 = k by 5, x 2 + y 2 = k 5 (1) 1 3 y=- x- 2 2 (2) is a tangent to (1). 9 = k 4 5 Ans:- (b) (2) Then c 2 = a 2 (m 2 +1) 1 9 = k.5 k = k = 9
54 Q26. If the circles x 2 + y 2 + 4x 4y - 6 = 0 and x 2 + y 2 +kx + 2y + 8 = 0 cut orthogonally then k = a) 3 b) -3 c) 10 d) -10
55 Solution:- Condition o is 2g 1 g 2 + 2f 1 f 2 = c 1 +c 2 i.e 2(2) k +2(-2).1= k 4 = 2 2k = 6 k = 3 Ans:- (a)
56 Q27. The radical centre of the circles c x 2 + y 2 = 5, x 2 + y 2-3x +1 = 0 and x 2 + y 2 +2y 1 =0 is a) (-2, 2), b) (2, 2), c) )(2, -2) d) )(1, 2)
57 Solution: Radical axis of x 2 + y 2 5 = 0 and x 2 + y 2 3x + 1 = 0 is 3x 6 = 0 x - 2 = 0 x = 2 Radical axis of x 2 + y 2-3x + 1 = 0 and x 2 + y 2 + 2y 1 = 0 is -3x 2y + 2 = 0 Substituting x = 2 we get = 2y 2y = -4 y = -2 Radical center = (2, -2) Ans:- (c)
58 Q28. The number of common tangents to the circles x 2 + y 2 + 2x + 8y 23 = 0 and x 2 + y 2-4x 10y + 19 = 0 are a) 1 b) 2 c) 3 d) 4
59 Solution:- C 1 = (-1, -4), C 2 = (2, 5), r 1 = 40 = 2 10, r 2 = 10 CC= 9+81= 90=3 10 = r + r Circles touch each other externally. There are 3 common tangents Ans:- (c)
60 Q29. The points from which the tangents to the circlesx 2 + y 2 8x + 40 = 0, 5x 2 + 5y 2 25x + 80 = 0 and x 2 + y 2 8x + 16y = 0 are equal in length is
61 a) 15 8, b) 15-8, 2 2 c) 15 8,- d) -8,
62 Solution :- The required point is radical centre of 3 circles. Radical axis of given circles are 3x 24 = 0 and 16y = 0 x = 8 and y = -15 The point = Ans:- (c) 8,
63 Q30. Two circles of equal radius r cut orthogonally. If their centres are (2, 3) and (5, 6) then r = a) 1 b) 2 c) 3 d) 4
64 Solution:- C 1 = (2, 3) C 2 = (5, 6) Eqn x 2 + y 2 4x 6y + 13 r 2 = 0 (1) x 2 + y 2 10x 12y + 41 r 2 = 0 (2) (1) cuts (2) orthogonally 2g 1 g 2 + 2f 1 f 2 = c 1 + c 2 2(2) 5 + 2(3) 6 = 13 r r = 74 2r 2 2r 2 = = 18 r 2 = 9 r = 3 Ans:- (c)
65 r r (2, 3) (5, 6) OR 2 2r 2 = ( ) 2 2 (2-5) +(3-6) 2r 2 = ( ) = 18 r 2 = 9 r = 3 Ans:- (c)
66 Q31. The number of tangents which can be drawn from the point (1, 2) to the circle x 2 + y 2 2x 4y + 4 = 0 are a) 1 b) 2 c) 3 d) 0
67 Solution: Power of (1, 2) w.r.t. the circle = = -1 is negative Point lies inside the circle Number of tangents that can be drawn to the circle from (1, 2) = 0 Ans:- (d)
68 Q32. The equation of the tangent to the circle x 2 + y 2 = 9 which are parallel to the line 2x + y 3 = 0 is a) )y = 2x ± 3 5 b) )y = -2x ± 3 5 c) y = -2x d) X + 2y = 0
69 Solution: Let the equation of the tangent be y = mx+c Since the tangent is parallel l to the line 2x + y 3 = 0, the slope of the tangent = -2
70 m = -2 and radius is r = 3 But c 2 = r 2 (m 2 + 1) c 2 = 32 [(-2) 2 + 1] = 45 c = ± 45 = ± 3 5 equation of tangent is y = mx + c = -2x ± 3 5 Ans: (b)
71 Q33. The length of the tangent from (3, -4) to the circle 2x 2 + 2y 2 7x 9y 13 = 0 is a) 26 b) 26 c) 2 d) 6
72 Solution:- Circle is x 2 + y x - y - = length of tangent from (3, -4) = ( 4) (3) 9 ( 4) = = = 26 Ans:- (a)
73 Q34. The locus of the centre of a circle of radius 2 which rolls on the outside of the circle x 2 + y 2 +3x 6y 9 = 0 is a) x 2 + y 2 + 3x 6y + 5 = 0 b) x 2 + y 2 + 3x 6y - 31 = 0 c) x 2 + y 2 + 3x 6y + 29 = 0 4 d) x 2 + y 2 + 3x + 6y + 29 = 0
74 Solution:- Centre of given circle = = 4 4 r = 9 3, = 2 9 = BC AB = 2. The locus of centre = C of outside circle is also a circle concentric with given circle A B C
75 Its radius 9 13 = AC = BA + BC = 2 + = 2 2 Its equation is 3 x (y 3) 2 = x 2 + y 2 + 3x 6y 31 = 0 Ans: - (b)
76 Q35. The area of the triangle formed by the tangent at the point (a, b) to the circle x 2 + y 2 = r 2 and the coordinate axis is r 4 a) b) 2ab 2 r 4 ab r 4 d) c) ab ab r 4
77 Solution :- Tangent at (a, b) to the circle x 2 + y 2 = r 2 is ax + by = r 2 that is same as + ( r ) ( ) 2 r This meets x-axis at A and y-axis at B 0, 2 r b r 2, 0 a 1 Area of OAB = (OA) (OB) = 1 r Ans:- (b) x a y 2 = 1 b ab
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