Q1. If (1, 2) lies on the circle 2 + 2 + 2 + 2f + = 0 x2 + y2 + 2gx + 2fy + c = 0 which is concentric with the circle x2 + y2 +4x + 2y 5 = 0 then c =

a) 11 b) 13 c) 24 d) 100 a) 11 b) -13 c) 24 d) 100

Solution: Solution: Any circle concentric with x2 + y2 + 4x + 2y 5 = 0 is x y 4x 2y 5 0 is x2 + y2 + 4x + 2y + c = 0. S b tit ti (1 2) Substituting (1, 2), 1 + 4 + 4 + 4 + c = 0 c= -13 Ans: (b) Ans: (b)

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 q a) x2 + y2 - 6x + 2y 28 = 0 a) x + y - 6x + 2y 28 = 0 b) x2 + y2 6x + 2y + 28 = 0

) 2 2 4 2 24 0 c) x2 + y2 + 4x 2y + 24 = 0 d) x2 + y2 + 2x 2y - 12=0

Solution: Solution: =AB 6 , OC perpendicular distance

f (3 1) t 2 5 18 0 from (3, -1) to2x 5y + 18 = 0 + +

= =+

6 5 18 294 25

A BC+4 25

= = + = + =2 2

r OB OC BC 29 9 38 ( 3)2 ( 1)2 38

A BC

O (3 1) (x 3)2 + (y + 1)2 = 38 x2 + y2 6x + 2y 28 = 0

O (3, -1)

Ans: (a)Ans: (a)

Q3.The four distinct points, (0, 0), (2, 0), (0, -2) and (k, 2) are concyclic if k = concyclic if k =

a) 2 b) 2 c) 0 d) 1a) 2 b) -2 c) 0 d) 1

S l ti Solution: (2, 0) and (0, -2) are ends of (2, 0) and (0, 2) are ends of

the diameter Eqn is x2 + y2 ax by = 0 x2 + y2 2x + 2y = 0 x + y 2x + 2y = 0 Substituting (k, -2) , k2 + 4 2k

4 = 0 k2 2k = 0, k =2 (k 0) Ans: (a)Ans: (a)

Q4. The length of the intercept made by the circle

x2 + y2 + 10x 12y 13 = 0 x + y + 10x 12y 13 = 0 on the y axis is

) 1 b) 2 ) 3 d) 14a) 1 b) 2 c) 3 d) 14

Solution: Length =

22 f - c =2 36 +13 =14 Ans: (d)

Q5. The length of the chord x = 3y + 13 of the

i l 2 + 2 4 + 4 + circle x2 + y2 4x + 4y + 3 = 0 is 3 = 0 is

a) 2 5 b) 5 2 a) 2 5 b) 5 2 c) 3 5 d) 10) )

Solution: OB= radius of the circle =

BA C

4+4-3 = 5 OC = perpendicular distance

O (2,-2)

OC = perpendicular distance from (2, -2) to x - 3y 13 = 0

2+6-13 5= =1+9 10

2 2 25 10 2 2 25 10AB=2BC =2 OB - OC =2 5 - = = 1010 10

Ans: (d) Ans: (d)

Q6. The area of the circle whose equations are given by x = 3 + 2 cos and y = 1 by x 3 + 2 cos and y 1 + 2 sin is

a) 4 b) 6 c) 9 d) 8 a) 4 b) 6 c) 9 d) 8

Solution: Gi 3 2 d Given x - 3 = 2cos and

y 1 = 2 sin y 1 = 2 sin. (x 3)2 + (y 1)2 = 4 (x 3) + (y 1) = 4 radius = 2 Area = 4 radius = 2 Area = 4

Ans: (a) Ans: (a)

Q7. A circle having area 9sq units touches both thesq. units touches both thecoordinate axes in the thirdquadrant, then its equation is

a) x2 + y2 6x 6y 9 =0 a) x2 + y2 6x 6y 9 =0 b) x2 + y2 + 6x + 6y + 9 =0 ) y y c) x2 + y2 6x + 6y + 9 = 0

2 2 d) x2 + y2 + 6x 6y + 9 =0

Solution: Solution: Area = 9 r2 = 9 r2 = 9 r = 3Since circle touches the coordinateaxes in the third quadrant, itsaxes in the third quadrant, itscentre must be (-3, -3)

Eqn is (x h)2 + (y k)2 = r2 Eqn is (x h)2 + (y k)2 = r2 (x + 3)2 + (y + 3)2 = 9 x2 + y2 + 6x + 6y + 9 = 0 Ans: (b) Ans: (b)

Q8. If x + y + k = 0 is a tangentQ8. If x y k 0 is a tangentto the circle

x2 + y2 2x 4y + 3 = 0 then k =

a) 20 b) -1,-5 c) 2 d) 4 a) 20 b) 1, 5 c) 2 d) 4

Solution:- (1, 2)

x + y + k = 0

radius = 1+ 4 - 3 = 2 Perpendicular distance from (1, 2)

y

Perpendicular distance from (1, 2) to x + y + k = 0} = radius

2k21 ++ = 2 K + 3 = 2 k = -1, -5

A (b) Ans(b)

Q9 Th di f Q9. The radius of any circle touching the lines any circle touching the lines 3x - 4y + 5 = 0 and 3x 4y 5 0 and 6x 8y 9 =0 is a) 1 b) 2315 c) 2019 d) 1920

Solution: 3x 4y + 5 = 0 ..1

6x 8y 9 = 0 2 6x 8y 9 = 0 2 Multiply 1 by 2 6x 8y + 10 = 0

Given lines are parallel lines p 2r = distance between the lines

2 2c - d 10 + 9 192r= = = 10a + b 36 +64

r = 1920

Ans: (d)

Q10. The circles x2 + y2 10x + 16 = 0 and x2 + y2 10x + 16 = 0 and x2+ y2 = r2 intersect each x + y r intersect each other in distinct points if p

a) r > 8 b) r < 2 c) 2 < r < 8 d) 2 r 8

S l ti Solution: C1 = (5, 0), r1 = 3, C2 = (0, 0), r2 = r

Circles intersect if r2 r1 < c1c2 < r1 + r2

r 3 < c1c2 < 3 + r r 3 < 5 < 3 + r r 3 < 5 r < 8 r + 3 > 5 r > 2

2 < r < 8

Ans: (c)

Q11. If ax2 + by2 + (a + b 4) xy ax by 20 = 0 If ax + by + (a + b 4) xy ax by 20 = 0

represents a circle, then its radius is

a) 21 b) 42 a) 2

b) 2

c) 2 21 d) 22 ) )

Solution: Solution: Since the equation represents a circle. Coefficient of x2 and y2 are equal a = b Also xy coefficient = 0 a + b 4 = 0 2a 4 = 0 a = 2 b = 2 equation is 2x2 + 2y2 2x 2y 20 = 0 equation is 2x 2y 2x 2y 20 0 i.e. x2 + y2 x y 10 = 0

= 1 1 42+ +10 = r = 1 1 42+ +10 =4 4 2

Ans: (b) Ans: (b)

Q12 If the centroid of an 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 the equation of the circumcircle is

a) x2 + y2 - 2x - 2y - 3 = 0 ) y y b) x2 + y2 + 2x 2y - 3 = 0 c) x2 + y2 + 2x + 2y - 3 = 0 d) 2 + 2 2 + 3 3 0 d) x2 + y2 - 2x + 3y - 3 = 0

Solution: Centroid = circumcentre

centre = (1 1) centre = (1, 1)

Radius = ( ) ( )2 21+ 1 + 1 - 2 = 5 Radius ( ) ( )1 1 1 2 5

(x 1)2 + (y 1)2 = ( )25 ( ) (y ) ( ) x2 + y2 2x 2y + 2 5 = 0

2 2 2 2 3 0 x2 + y2 2x 2y 3 = 0 Ans: (a)

Q13. The lines ax + by + c = 0, bx + cy + a = 0 and bx cy a 0 and cx + ay + b =0 are concurrent if a) 2a = 3abc b) a =0 a) 2a = 3abc b) a =0 c) 2a = ab d) 2a =0 ) )

Solution: Solution: a b cb c a =0 c a b

a (bc a2) b (b2 ca) ( ) ( ) + c (ab c2) = 0 abc a3 b3 + abc + abc c3 = 0 abc a3 b3 + abc + abc c3 = 0 a3 + b3 + c3 = 3abc a + b + c = 0 a = 0 Ans: (b) Ans: (b)

Q14 The area of the Q14. The area of thetriangle whose sides arealong the lines x = 0, y = 0and 4x + 5y 20 = 0 is and 4x + 5y 20 = 0 is

a) 20 sq. units b) 10 sq. units c) 1/10 sq. units c) 1/10 sq. units d) 1/20 sq. unit

Solution: Solution:

x = 0

4 + 5 = 20

B (0, 4)

A (5, 0)

4x + 5y = 20

A f OAB 1

O y = 0

( , )

Are of OAB = 1 OA.OB2

=1 5.4 =2 10 Ans (b) 2 0 s (b)

Q15 The line x + y = 4 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

Solution: Take A= (-1, 1) and B = (5, 7) Let x + y 4 =0 divide AB in the ratio k : 1

2 1 2 1mx +nx my +ny 5k-1 7k+1P= , = ,m+n m+n k+1 k+1

Substituting in x + y 4 = 0, we get 5k-1 7k+1+ -4=0k+1 k+1 k+1 k+1 5k 1+7k+1-4k-4=0 8k = 4 1k=2 the ratio is k:1 = 1:2 Ans: (b)

Or

b 1 1

2 2

- ax +by +cm=n ax +by +c

m -1+1-4 4 1=- = =n 5+7 4 8 2 n 5+7-4 8 2

m:n = 1:2 Ans: (b)

Q16. If p is the length of the perpendicular from the origin perpendicular from the origin on the line whose intercepts on the axes are a and b, then 1 = then 1 =2p

a) 1 1+ b) 1 1 a) 2 2+a b b) 2 2.a b c) 2 21 1-a b d) 22

1 1-ab a b ab

Solution:

(a 0)

(0, b)

Equation is x + by = 1

(a, 0)

qa b

Length 1 12 2

c -1= p=+b

g1 12 2

2 2a +ba +b

2 2 2 22

2 2ab a b 1 a +b 1 1p= p = = = +b b b 2 22 2 2 2 22 2p p p aa +b a b ba +b

Ans: (a) ( )

Q17. The reflection of the point (4, 4) with respect to the line x + y = 2 is to the line x + y = 2 is a) (1, 1) b) (- 2, - 2) ) 1 1 d) ( 2 2) c)

1 1- , -2 2

d) (-2, - 2)

Solution: (4, 4)

( , )

(h k)x + y 2

Mid point (4,4) & (h, k) =

h+4 k+4,2 2

S b tit ti i 2 0 t

(h, k)

Substituting in x + y 2 = 0, we get h+4 k+4+ -2=02 2 h = k .. (1) Also k-4-1=-1h-4 h + 4 + k + 4 - 4 = 0 h + k = -4 (2) h + k 4 (2) Solving (1) & (2), we get h=-2, k= -2 Ans. (d) Ans. (d)

Q18 Th i t f Q18. The point of i t ti f i f liintersection of pair of lines

2x2 5xy 3y2 + 6x +17y 20 = 0 is 2x2 5xy 3y2 + 6x +17y 20 = 0 is

a) (2, 1) b) (1, 2) ) ( , ) ) ( , )c) (1, -2) d) (-2, 1)

Solution:- Point of intersection Point of intersection

= 2 2hf -bg gh-af,b h b h = 2 2

,ab-h ab-h

-5 17-5 17 3 -2 +3(3) 2 2 -15-34-85+36 =

3 (3) 2 22 2 ,24 25-6- -6-4 4 =

-15-34-85+36 ,-24 -25 -24-25

=

-492-49,-49 -49 = (1, 2

Or Using partial derivative method, Difft. W.r.to x , keeping y Difft. W.r.to x , keeping y

constant 4x 5y +6 = 0 Difft. W. r. to y, keeping x constant y, p g -5x 6y + 17 = 0 S l i 1 d 2 Solving x = 1, and y = 2 Ans:- (b)

Q19 f fQ19. If the point of intersection of the lines intersection of the lines

kx+4y+2 = 0, x3y+5 = 0 y , y lie on 2x+7y 3 =0, then k= a) 2 b) 3 c) -2 d) -3

Solution :- Solution :- Solving, Substituting x 3y + 5 = 0 and

x= -2 and y =1 in kx + 4y +2 =0,

2x + 7y 3 = 0 We get

we get, We get, y = 1, x = -2 -2k + 4 + 2 = 0

k = 3 Ans:- (b) Ans: (b)

Q20 The equation of the line Q20. The equation of the line perpendicular to 5x 2y -7 = 0 and

passing through the point of intersection of the lines intersection of the lines

2x + 3y 1 = 0 and 3x + 4y 6 =0 is

a) 2x + 5y + 17 = 0 b) 2x + 5y 17 = 0 ) y c) 2x -5y + 17 = 0 d) 2 5 17 0 d) 2x 5y 17 = 0

Solution:- Solving, 2x + 3y = 1 3x + 4y = 6 ------------------- x = 14, y = -9 , y

Line perpendicular to p p 5x 2y 7 = 0 is 2x + 5y + k = 0 Substituting (14, -9 ) We get 28 45+k = 0 k = 17 28- 45+k = 0 k = 17 Equation is 2x+5y+17 = 0 q yAns:- (a)

Q21 If ( 4 5) is one vertex 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 equation of second diagonal is a) x + 3y 21 = 0 b) 2x - 3y - 7 = 0 ) 7 31 0 c) x + 7y - 31 = 0 d) 2x + 3y - 21 = 0 d) 2x + 3y - 21 = 0

S l ti Solution: The second diagonal passes g p through (-4, 5) and perpendicular to 7x y + 8 = 0 to 7x y + 8 = 0 Equation is x + 7y +k =0 Substituting (-4, 5) , -4 + 35 + k = 0 k = -31 Equation is x+ 7y - 31 = 0. Ans: (c) Ans:- (c)

Q22 The distance Q22. The distance between the pair of between the pair of parallel lines parallel lines

9x2 6xy + y2 + 18x 6y + 8 = 0 is y y y a) 1

10 b) 2

10

10 10

c) 410

d) 10 10

Solution:- distance =

=+

2 81 722 2 9 10g aca a b

= =1 2210 10

Ans:- (b)

Q23 If th f th Q23. If the sum of the slopes of the lines given by slopes of the lines given by x2 2k xy 7y2 = 0 is four y y times their product then k = ) 1 b) 1 ) 2 d) 2 a) 1 b)-1 c) 2 d) -2

S l ti Solution:-

+ 4 m1 + m2 = 4. m1m2

2h 4 a 2h 4 b2h = 4 . b

a - 2h = 4a

-2(-k) = 4 k = 2

Ans:- (c)

Q24. The area of the largest square inscribed in the circle square inscribed in the circle

x2 + y2 6x 8y = 0 is y y

) 100 it b) 25 it a) 100 sq units b) 25 sq units

) 10 it d) 50 it c) 10 sq units d) 50 sq units

Solution: Solution:- g = -3., f = -4, c = 0 g , ,

Radius = 9+16 = 5 5

a2 + a2 = 102 2a2 = 100 a2 = 50

a a 5

2a2 = 100 a2 = 50 Area = 50 sq units q

Ans:- (d)

Q25. If 2y + x + 3 = 0 is a tangent to 5x2 + 5y2 = k tangent to 5x2 + 5y2 = k, then the value of K is then the value of K is

a) 4 b) 9

c) 16 d) 25

Solution :- 5x2 + 5y2 = k

by 5, x2+ y2 = k5 (1)

1 3y = - x -2 2 -------- (2)

(2) is a tangent to (1) (2) is a tangent to (1). Then c2= a2 (m2+1) 4

9 = 5k

+14

1 49 = 5.4

k.5 k = 9

k = 9 Ans:- (b) Ans: (b)

Q26 If the circles Q26. If the circles x2 + y2 + 4x 4y - 6 = 0 and x + y + 4x 4y 6 0 and x2 + y2 +kx + 2y + 8 = 0 cut orthogonally then k = ) 3 b) 3 a) 3 b) -3 c) 10 d) -10 ) )

Solution: Solution:- Condition is Co d t o s 2g1 g2 + 2f1f2 = c1+c2 i.e

k2(2) +2(-2).1=-6+82

2k 4 = 2 2k = 6 k = 3 2k = 6 k = 3

Ans:- (a) Ans: (a)

Q27 The radical centre of Q27. The radical centre of the circles x2 + y2 = 5, t e c c es y 5,x2 + y2- 3x +1 = 0 and x2 + y2 +2y 1 =0 is a) (-2 2) b) (2 2) a) (-2, 2), b) (2, 2), c) (2, -2) d) (1, 2) ) ( , ) ) ( , )

Solution: Solution: Radical axis of x2+ y2 5 = 0 and

2+ 2 3 + 1 0 i 3 6 0 x2+ y2 3x + 1 = 0 is 3x 6 = 0 x - 2 = 0 x = 2

Radical axis of x2 + y2 - 3x + 1 = 0 and x2 + y2 + 2y 1 = 0 is -3x 2y + 2 = 0 Substituting x = 2 we get -6 + 2 = 2y 2y = -4 y = -2 g y y y Radical center = (2, -2)

Ans:- (c) Ans:- (c)

Q28. The number of common tangents to the circles tangents to the circles

x2 + y2 + 2x + 8y 23 = 0 and

x2 + y2 - 4x 10y + 19 = 0 are

a) 1 b) 2 a) 1 b) 2 c) 3 d) 4

Solution:- Solution:- C1 = (-1, -4), C2 = (2, 5), r1 = 40 =2 10 , r2 = 10 1 2C C = 9+81= 90=3 10 = r1 + r2 Circles touch each other Circles touch each other externally. There are 3 common tangents

Ans: (c) Ans:- (c)

Q29. The points from which the tangents to the circlesx2 + y2 8x + 40 = 0 circlesx + y 8x + 40 = 0, 5x2 + 5y2 25x + 80 = 0 andx2 + y2 8x + 16y + 160 = 0

l i l th iare equal in length is

15 15a)

158,2

b)

15-8,2

15 15c)

158,-2

d)

-15-8, 2

Solution :- Solution :- The required point is radical centre of 3 circles. Radical axis of given circles are 3x 24 = 0 and 16y + 120 = 0 3x 24 0 and 16y + 120 0 x = 8 and y = -15

2

The point =

-158,2

Ans: (c)Ans:- (c)

Q30 Two circles of equal Q30. Two circles of equal radius r cut orthogonally. radius r cut orthogonally.

If their centres are (2, 3) and (5, 6) then r = ) 1 b) 2 a) 1 b) 2 c) 3 d) 4 c) 3 d)

Solution:- C1 = (2, 3) C2 = (5, 6) Eqn x2 + y2 4x 6y + 13 r2 = 0 (1) q y y ( )

x2 + y2 10x 12y + 41 r2 = 0 (2) (1) cuts (2) orthogonally (1) cuts (2) orthogonally 2g1 g2 + 2f1f2 = c1+ c2 2(2) 5 + 2(3) 6 = 13 r2 + 61 r2 2(2) 5 + 2(3) 6 = 13 r2 + 61 r2 20 + 36 = 74 2r2 2r2 = 74 56 = 18 r2 = 9 r = 3 Ans:- (c)

r r

r r

(2, 3) (5, 6)

OR

2 2 ( )22 2 2r2 = ( )2 2(2 - 5) +(3 -6) 2r2 = ( )29+9 = 18 2r ( )9 9 18 r2 = 9 r = 3 Ans:- (c)

Q31 The number of Q31. The number of tangents which can be drawn from the point

(1 2) t th i l (1, 2) to the circle x2 + y2 2x 4y + 4 = 0 are y y 0 a e a) 1 b) 2 c) 3 d) 0 c) 3 d) 0

Solution: Solution: Power of (1, 2) w .r .t. the

circle = 1 + 4 2 8 + 4 = -1

is negative Point lies inside the circle Number of tangents that

can be drawn to the circle can be drawn to the circle from (1, 2) = 0 Ans: (d) Ans:- (d)

Q32 The equation of the Q32. The equation of the tangent to the circle tangent to the circle

x2 + y2 = 9 which are parallel x + y = 9 which are parallel to the line 2x + y 3 = 0 is to the line 2x y 3 0 is

a) y = 2x 3 5 b) y = -2x 3 5 ) y ) y c) y = -2x d) X + 2y = 0

S l ti Solution: Let the equation of the Let the equation of the

tangent be y = mx+c Since the tangent is ll l t th li parallel to the line 2x + y 3 = 0, the slope of 2x y 3 0, the slope of

the tangent = -2

m = 2 and radius is r = 3 m = -2 and radius is r = 3 But c2 = r2(m2 + 1)

c2 = 32 [(-2)2 + 1] = 45 45 3 5 c = 45 = 3 5 equation of tangent is q g y = mx + c = -2x 3 5 Ans: (b)( )

Q33 The length of the Q33. The length of the tangent from (3, -4) g ( , )

to the circle 2x2 + 2y2 7x 9y 13 = 0 is

a) 26 b) 26 a) 26 b) 26

c) 2 d) 6 c) 2 d) 6

Solution:- Solution:

Circle is x2 + y2 - 27x -

29y -

213 = 0

2 2 2

length of tangent from (3, -4)

= 2

13)4(

29

)3(27

)4(3 22 +

= 2

1318

221

169 ++ = 1743

= 26

Ans (a)Ans:- (a)

Q34 The locus of the centre Q34. The locus of the centre of a circle of radius 2 which rolls on the outside of the circle x2 + y2+3x6y9 = 0 is circle x + y +3x 6y 9 0 is

a) x2 + y2 + 3x 6y + 5 = 0 b) x2 + y2 + 3x 6y - 31 = 0 ) 2 + 2 + 3 6 + 29 0 c) x2 + y2 + 3x 6y +

429 = 0

d) x2 + y2 + 3x + 6y + 29 = 0 d) x + y + 3x + 6y + 29 = 0

Solution:- Centre of given circle =

3,

23 = C

9 81 9A

r = 9949

++ = 481 =

29 = BC

AB = 2 C

AB

AB = 2. The locus of centre of outside circle is also a circle

concentric with given circle g

Its radius

= AC = BA + BC = 2 + 29 =

213

Its equation is

2

23

x

+ (y 3)2 =

2

213

x2 + y2 + 3x 6y 31 = 0

Ans: - (b)

Q35 Th f th Q35. The area of the triangle formed by the g ytangent at the point (a, b) to the circle x2 + y2 = r2 to the circle x2 + y2 = r2 and the coordinate axis is

a) ab2r4 b)

|ab|2r4

||

c) abr4 d)

|ab|r4

ab |ab|

Solution :- Tangent at (a, b) to the circle x2 + y2 = r2 isg ( , ) y

ax + by = r2 that is same as ( )arx2

+ ( )bry2

= 1

This meets x-axis at A

0,ar2

2and y-axis at B

br

,02

Area of OAB = 1 (OA) (OB) = 1 r4

Area of OAB = 21 (OA) (OB) =

21

|ab|r

Ans:- (b) ( )