Q1. If (1, 2) lies on the circle
x2 + y2 + 2gx
2
+ 2fy
2f + c = 0
which is concentric with
the circle x2 + y 2 +4x + 2y –
5 = 0 then c =
a) 11
b) -13
13
c) 24
d) 100
Solution:
Any circle concentric with
x + y + 4x + 2y – 5 = 0 is
x2 + y 2 + 4x + 2y + c = 0.
2
2
Substituting
S
b tit ti
(1,
(1 2)
2),
1+ 4+ 4 + 4 + c = 0
⇒ c= -13
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
q
of the circle is
a)
b)
c))
d)
x2 +
x2 +
x2 +
x2 +
y2 - 6x + 2y – 28 = 0
y2 – 6x + 2y + 28 = 0
y2 + 4x
4 – 2y
2 + 24 = 0
y2 + 2x – 2y - 12=0
Solution:
AB = 6 , OC perpendicular distance
f
from
(3
(3, -1)
1) to2x
t 2 – 5y
5 + 18 = 0
=
6 + 5 + 18
4 + 25
= 29
2
A
C
2
r = OB = OC + BC = 29 + 9 = 38
∴ (x
( – 3)2 + (y
( + 1)2 = 38
⇒ x2 + y2 – 6x + 2y – 28 = 0
Ans: (a)
O (3,
(3 -1)
1)
B
Q3.The four distinct points, (0,
0), (2, 0), (0, -2) and (k, –2) are
concyclic if k =
a) 2
b) -2
2
c) 0
d) 1
S l ti
Solution:
(2, 0) and (0, -2)
2) are ends of
the diameter
∴ Eqn is x2 + y2 – ax – by = 0
⇒ x + y – 2x + 2y = 0
2
2
Substituting (k, -2) , k + 4 – 2k –
2
4 = 0 k – 2k = 0, k =2 (k ≠ 0)
2
Ans: (a)
Q4. The length of the
intercept made by the circle
2
2
x + y + 10x – 12y – 13 = 0
on the y – axis is
a)) 1
b) 2
c)) 3
d) 14
Solution:
Length =
2 f 2 - c = 2 36 + 13 = 14
Ans: (d)
Q5. The length of the
chord x = 3y + 13 of the
2
2
circle
i l x + y – 4x
4 +4
4y +
3 = 0 is
a) 2 5
b) 5 2
c)) 3 5
d)) 10
Solution:
A
OB= radius of the circle =
C
O (2,-2)
4+4-3 = 5
= perpendicular distance
from (2, -2) to x - 3y – 13 = 0
OC
= 2+6-13 = 5
1+9
10
AB = 2BC = 2 OB - OC = 2 5 - 25 = 10 = 10
10
10
2
Ans: (d)
2
B
Q6. The area of the circle
whose equations are given
by x = 3 + 2 cos θ and y = 1
+ 2 sinθ is
a) 4π
b) 6π
c) 9π d) 8π
Solution:
Gi
Given
x-3=2
2cosθ
θ and
d
y – 1 = 2 sinθ.
sinθ
2
2
⇒ (x – 3) + (y – 1) = 4
⇒ radius = 2 Area = 4π
∴ Ans: (a)
Q7. A circle having area 9π
sq units touches both the
sq.
coordinate axes in the third
quadrant, then its equation is
2
2
a) x + y – 6x – 6y – 9 =0
2
2
b)) x + y + 6x + 6y
y + 9 =0
2
2
c) x + y – 6x + 6y + 9 = 0
d) x2 + y2 + 6x – 6y + 9 =0
Solution:
Area = 9π ⇒ πr2 = 9π ⇒ r2 = 9 ⇒ r = 3
Since circle touches the coordinate
axes in the third quadrant, its
centre must be (-3, -3)
∴ Eqn is (x –h)
h) + (y – k) = r
2
2
⇒ (x + 3)2 + (y + 3)2 = 9
2
2
⇒ x + y + 6x + 6y + 9 = 0
Ans: (b)
2
Q8. If x + y + k = 0 is a tangent
to the circle
x + y – 2x – 4y + 3 = 0 then k =
2
2
a) ± 20 b) -1,-5
1, 5
c) ± 2
d) 4
Solution:(1, 2)
radius =
x+y+k=0
1+ 4 - 3 = 2
Perpendicular distance from (1, 2)
to x + y + k = 0} = radius
1+ 2 + k
2
=
A (b)
Ans(b)
2
⇒ K + 3 = ±2 ⇒ k = -1, -5
Q9 The
Q9.
Th radius
di
off
any circle touching the lines
3x - 4y + 5 = 0 and
6x – 8y – 9 =0 is
a) 1
b)
23
15
c)
20
19
d)
19
20
Solution:
3x – 4y + 5 = 0 …………..1
6x – 8y – 9 = 0 ………………2
Multiply 1 by 2
6x – 8y + 10 = 0
Given lines are p
parallel lines
∴ 2r = distance between the lines
2r=
c - d = 10 + 9 = 19 ⇒ r = 19
20
10
a2 + b2
36 + 64
Ans: (d)
Q10. The circles
x + y – 10x + 16 = 0 and
2
2
x + y = r intersect each
2
2
2
other in distinct p
points if
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
represents a circle, then its
radius is
21
a)
2
42
b)
2
c)) 2 21
d)) 22
Solution:
Since the equation represents a circle.
Coefficient of x2 and y2 are equal ∴ a = b
Also xy – coefficient = 0
∴ 2a – 4 = 0
∴a=2
∴a+b–4=0
∴b=2
∴ equation is 2x2 + 2y2 – 2x – 2y – 20 = 0
i.e. x2 + y2 – x – y – 10 = 0
∴ r = 1 + 1 +10 = 42
4 4
2
Ans: (b)
Q12 If the centroid of an
Q12.
equilateral triangle is (1, 1) and
one of the vertex is (-1, 2). Then
the equation of the circumcircle is
a)) x2 + y2 - 2x - 2y
y-3=0
b) x2 + y2 + 2x – 2y - 3 = 0
c) x2 + y2 + 2x + 2y - 3 = 0
d) x2 + y2 - 2x
2 +3
3y - 3 = 0
Solution:
Centroid = circumcentre
∴ centre = (1
(1, 1)
Radius =
2
2
(1 + 1) + (1 - 2 ) = 5
∴ ((x – 1)) + (y – 1)) =
2
2
( 5)
2
⇒ x2 + y2 – 2x – 2y + 2 – 5 = 0
x2 + y2 – 2x
2 – 2y
2 –3=0
Ans: (a)
Q13. The lines ax + by + c = 0,
bx + cy + a = 0 and
cx + ay + b =0 are concurrent if
a)
c))
∑ a2 = 3abc
∑ a2 = ∑ ab
b)
d))
∑ a =0
∑ a2 =0
Solution:
a
b
b
c
c
a =0
c
a
b
⇒ a ((bc – a2) – b ((b2 – ca))
+ c (ab – c2) =
0
⇒ abc – a3 – b3 + abc + abc – c3 = 0
⇒ a3 + b3 + c3 = 3abc ⇒ a + b + c = 0
⇒Σa=0
Ans: (b)
Q14. The area of the
Q14
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
Solution:
x=0
B (0, 4)
4 +5
4x
5y = 20
A ((5,, 0))
O
A off ∆OAB =
Are
= 21 5.4 = 10
0
y=0
1 OA.OB
2
∴ Ans
s (b)
Q15. The line x + y = 4
Q15
divides the line joining
the points (-1, 1)
and (5, 7) in the ratio
a)) 2 : 3
c) 1 : 1
b)) 1 : 2
d) 4 : 3
Solution:
Take A= (-1, 1) and B = (5, 7)
Let x + y – 4 =0 divide AB in the ratio k : 1
mx2 +nx1 my2 +ny1 ⎞⎟ ⎛⎜ 5k -1 7k+1⎞⎟
P=
,
=⎜
,
⎟
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
k= 1 ⇒ the ratio is k:1 = 1:2
2
Ans: (b)
Or
⎛
⎞
ax
+by
b
+c
⎜
⎟
1
m= ⎝ 1
⎠
n ax +by +c
2
2
m=- ⎡⎢ -1+1-4⎤⎥ = 4 = 1
n ⎢⎣ 5+7-4
5+7 4 ⎥⎦ 8 2
m:n = 1:2
Ans: (b)
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
a)
c)
1 =
p2
1+1
a2 b2
1-1
a2 b2
b)
d)
1.1
a2 b2
1-1
b2 a2
Solution:
(0, b)
(a 0)
(a,
x
y
Equation
q
is
+ =1
a b
Length
g
=
⇒p
p=
c
⇒ p=
a2 +b2
-1
1
2
a +b
1
2
ab ⇒p2 = a2b2 ⇒ 1 = a2 +b2 = 1 + 1
p2 a2b2 b2 a2
a2 +b
b2
a2 +b2
Ans: (a)
( )
Q17. The reflection of the
point (4, 4) with respect
to the line x + y = 2 is
a) (1, 1)
b)
c))
d) (-2,
( 2 - 2)
⎛
⎜
⎜⎜
⎝
⎞
1
1
⎟
,⎟
2
2 ⎟⎠
(- 2, - 2)
Solution:
((4,, 4))
x+ y–2
(h k)
(h,
Mid point (4,4) & (h, k) =
⎛
⎜
⎜
⎝
h+4 , k +4 ⎞⎟
2
2 ⎟⎠
S b tit ti
Substituting
in
i x+y–2=0
0, we gett
h+4 + k+4 -2=0 ⇒ h = k
2
2
Also kh-- 44 ×-1=-1 ⇒ h + 4
……….. (1)
+k+4-4=0
⇒ h + k = -4
4 …………… (2)
Solving (1) & (2), we get h=-2, k= -2
Ans. (d)
Q18 The
Q18.
Th point
i t off
i t
intersection
ti off pair
i off li
lines
2x – 5xy – 3y + 6x +17y – 20 = 0 is
2
a)) ((2,, 1))
c) (1, -2)
2
b)) ((1,, 2))
d) (-2, 1)
Solution:Point of intersection
hf -bg , gh-af ⎞⎟
= ab-h
b h2 ab-h
b h2 ⎟⎟⎠
⎛
⎜
⎜⎜
⎝
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
=
⎛
⎞⎞
⎛
⎞
-5
17
⎜
⎜
⎟
-5 ×17 +3×(3)
3 (3) 3⎜ 2 ⎟ -2⎜ 2 ⎟⎟ ⎟⎟
2 2
, ⎝ ⎠ ⎝ ⎠ ⎟⎟
-6- 24
-6- 25 ⎟⎟
⎟
4
4
⎠
⎞
-49×2
-49
= -49, -49 ⎟⎟⎟ = (1, 2
⎠
⎛
⎜
⎜⎜
⎝
-85+36 , -15-34 ⎞⎟
= -24 -25 -24-25 ⎟⎟
⎠
⎛
⎜
⎜⎜
⎝
Or Using partial derivative method,
Difft. W.r.to
x , keeping y
constant
⇒ 4x – 5y +6 = 0
p g x constant
Difft. W. r. to y, keeping
⇒ -5x – 6y + 17 = 0
S l i
Solving
x=1
1, and
dy=2
Ans:- (b)
Q19 Iff the point off
Q19.
intersection of the lines
kx+4y+2
y
= 0,, x–3y+5
y
=0
lie on 2x+7y – 3 =0, then k=
a) 2
b) 3
c) -2
d) -3
Solution :Solving,
x – 3y + 5 = 0
Substituting
x= -2 and y =1 in
and
kx + 4y +2 =0,
2x + 7y – 3 = 0
we get,
We get,
get
y = 1, x = -2
-2k + 4 + 2 = 0
⇒
Ans:- (b)
Ans:
k=3
Q20 The equation of the line
Q20.
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
y – 17 = 0
c) 2x -5y + 17 = 0
d) 2x
2 – 5y
5 – 17 = 0
Solution:Solving, 2x + 3y = 1
3x + 4y = 6
------------------x = 14,, y = -9
Line p
perpendicular
p
to
5x – 2y – 7 = 0 is
2x + 5y + k = 0
Substituting (14, -9 ) We get
28 45+k = 0 ⇒ k = 17
28∴ Equation
q
is 2x+5y+17
y
=0
Ans:- (a)
Q21. If (-4,
Q21
( 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
7 - 31 = 0
d) 2x + 3y - 21 = 0
S l ti
Solution:
The second diagonal
g
passes
p
through (-4, 5) and perpendicular
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:-
Q22 The distance
Q22.
between the pair of
parallel lines
9x2 – 6xy
y + y2 + 18x – 6y
y + 8 = 0 is
a) 1
b) 2
10
10
c) 4
d) 10
10
Solution:distance =
2 − ac
g
81− 72
=
2 ⎛
2
9×10
a ⎜⎝ a + b ⎞⎟⎠
= 2⋅ 1 =
10
Ans:- (b)
2
10
Q23 If the
Q23.
th sum off th
the
slopes of the lines given by
x – 2k xy
y – 7y
y = 0 is four
2
2
times their product then k =
a)) 1
b) 1
b)-1
c)) 2
d) -2
2
S l ti
Solution:m1 + m2 = 4.
4 m1m2
− 2h = 4 . a ⇒ - 2h = 4
4a
b
b
-2(-k) = 4 ⇒ k = 2
Ans:- (c)
Q24. The area of the largest
square inscribed in the circle
x + y – 6x – 8y
y = 0 is
2
2
a)) 100 sq units
it
b) 25 sq units
it
c)) 10 sq units
it
d) 50 sq units
it
Solution:Solution:
g = -3.,, f = -4,, c = 0
5 a
•
a
Radius =
9+16 = 5
∴ a2 + a2 = 102
⇒ 2a2 = 100 ⇒ a2 = 50
∴ Area = 50 sq
q units
∴Ans:- (d)
Q25. If 2y + x + 3 = 0 is a
tangent to 5x2 + 5y2 = k,
k
then the value of ‘K’
K is
a) 4
b) 9
c) 16 d) 25
Solution :- 5x2 + 5y2 = k
÷ by 5, x2+ y2 =
y = -1 x -3
2
2
k
5
…… (1)
-------- (2)
(2) is a tangent to (1).
(1)
Then c2= a2 (m2+1)
9
4
=
k ⎛⎜⎜ 1 +1⎞⎟⎟
5 ⎜⎝ 4 ⎟⎠
⇒
9
4
=
⇒k=9
Ans:- (b)
Ans:
k.5
5.4
⇒k=9
Q26 If the circles
Q26.
x2 + y2 + 4x – 4y - 6 = 0 and
x + y +kx + 2y + 8 = 0 cut
2
2
orthogonally then k =
a)) 3
c)) 10
b) -3
3
d)) -10
Solution:Solution:
Co d t o is
Condition
s
2g1 g2 + 2f1f2 = c1+c2
⎛
⎞
k
i.e 2(2) ⎜⎜ ⎟⎟ +2(-2).1=-6+8
⎝
2⎠
2k – 4 = 2
⇒ 2k = 6 ⇒ k = 3
∴ Ans:
Ans:- (a)
Q27 The radical centre of
Q27.
tthe
e circles
c c es x + y = 5,
2
2
‘x2 + y2- 3x +1 = 0 and
x2 + y2 +2y – 1 =0 is
a) (-2,
(-2 2)
2),
b) (2
(2, 2)
2),
c)) ((2,, -2))
d)) ((1,, 2))
Solution:
Radical axis of x2+ y2– 5 = 0 and
x2+ y2– 3x
3 + 1 = 0 iis 3
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 g
get
-6 + 2 = 2y
y ⇒ 2y
y = -4 ⇒ y = -2
∴ Radical center = (2, -2)
Ans:- (c)
Q28. The number of common
tangents to the circles
x + y + 2x + 8y – 23 = 0 and
2
2
x + y - 4x – 10y + 19 = 0 are
2
2
a) 1
b) 2
c) 3
d) 4
Solution:C1 = (-1, -4), C2 = (2, 5),
r1 = 40 = 2 10 , r2 = 10
C1 C2 = 9+81 = 90 =3 10
= r1 + r2
∴ Circles touch each other
externally.
⇒ There are ‘3’ common tangents
∴ Ans:Ans: (c)
Q29. The points from
which the tangents to the
2
2
circlesx + y – 8x + 40 = 0,
0
2
2
5x + 5y – 25x + 80 = 0 and
2
2
x + y – 8x + 16y + 160 = 0
are equall iin llength
th iis
⎛ 15 ⎞
a) ⎜ 8,
⎟
2 ⎠
⎝
15 ⎞
⎛
b) ⎜ -8,
⎟
2 ⎠
⎝
15 ⎞
⎛
c) ⎜ 8,⎟
2 ⎠
⎝
⎞
-15
15
⎟
d) -8,
2 ⎟⎠
⎛
⎜
⎜
⎝
Solution :The required point is
radical centre of 3 circles.
∴ Radical axis of given
circles are
3x – 24 = 0 and 16y + 120 = 0
-15
2
⎛ -15 ⎞
⎜ 8,
⎟
2
⎝
⎠
∴ x = 8 and y =
∴ The point =
Ans: (c)
Ans:-
Q30 Two circles of equal
Q30.
radius r cut orthogonally.
If their centres are (2, 3)
and (5, 6) then r =
a)) 1
b) 2
c) 3
d) 4
Solution:C1 = (2, 3)
∴ Eqn
q
C 2 = (5, 6)
x2 + y2 – 4x – 6y
y + 13 – r2 = 0 …(1)
( )
x2 + y2 – 10x – 12y + 41 – r2 = 0 … (2)
(1) cuts (2) orthogonally
∴2g1 g2 + 2f1f2 = c1+ c2
2(2) 5 + 2(3) 6 = 13 –rr2 + 61 – r2
20 + 36 = 74 – 2r2
2r2 = 74 – 56 = 18
⇒ r2 = 9 ⇒ r = 3
Ans:- (c)
r
•
•
(2, 3)
OR
2
2 =
2r
(
2r =
(
2
2
(2 - 5) + (3 - 6)
9+9
)
2
2
)
2
= 18
⇒ r2 = 9 ⇒ r = 3
Ans:- (c)
r
(5, 6)
Q31. The number of
Q31
tangents which can be
drawn from the point
(1 2) to
(1,
t th
the circle
i l
x + y – 2x – 4y
y + 4 = 0 are
a e
2
2
a) 1
b) 2
c) 3
d) 0
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
from (1, 2) = 0
Ans: (d)
Ans:-
Q32 The equation of the
Q32.
tangent to the circle
x + y = 9 which are parallel
2
2
to the line 2x + y – 3 = 0 is
a)) y = 2x ± 3
c) y = -2x
5
b)) y = -2x ± 3
d) X + 2y = 0
5
S l ti
Solution:
Let the equation of the
tangent be y = mx+c
Since the tangent is
parallel
ll l tto th
the line
li
2x + y – 3 = 0, the slope of
the tangent = -2
∴ m = -2
2 and radius is
r=3
But c2 = r2(m2 + 1)
⇒ c2 = 32 [(-2)2 + 1] = 45
⇒c=±
45 = ± 3 5
∴ equation
q
of tangent
g
is
y = mx + c
( )
Ans: (b)
= -2x ± 3 5
Q33. The length of the
Q33
tangent
g
from ((3,, -4))
to the circle
2x + 2y – 7x – 9y – 13 = 0 is
2
2
a) 26
b) 26
c) 2
d) 6
Solution:
Solution:7
9
13
Circle is x + y - x - y =0
2
2
2
2
2
length of tangent from (3, -4)
7
9
13
(3) − (−4) −
2
2
2
=
3 + (−4) −
=
13
21
9 + 16 −
+ 18 −
=
2
2
=
26
2
Ans (a)
Ans:-
2
43 − 17
Q34. The locus of the centre
Q34
of a circle of radius 2 which
rolls on the outside of the
circle x2 + y2+3x
+3x–6y–9
6y 9 = 0 is
a) x2 + y2 + 3x – 6y + 5 = 0
b) x2 + y2 + 3x – 6y - 31 = 0
c)) x + y + 3x
3 – 6y
6 +
2
2
29
4
=0
d) x + y + 3x + 6y + 29 = 0
2
2
Solution:Centre of given circle =
r=
9
+9+9
4
=
81
4
=
9
2
⎛ 3
⎞
−
,
3
⎜
⎟
⎝ 2
⎠
= BC
=C
A
B
C
AB = 2
2.
The locus of centre
of outside circle is also a circle
concentric with g
given circle
∴ Its radius
= AC = BA + BC = 2 +
9
2
=
Its equation is
2
⎛
⎛ 13 ⎞
⎛ 3 ⎞⎞
2
+
(y
–
3)
=
x
−
−
⎜
⎜
⎟
⎟⎟
⎜
⎝ 2 ⎠
⎝ 2 ⎠⎠
⎝
x2 + y2 + 3x – 6y – 31 = 0
Ans: - (b)
2
13
2
Q35. The
Q35
Th area off th
the
triangle
g formed by
y the
tangent at the point (a, b)
to the circle x2 + y2 = r2
and the coordinate axis is
a)
r
2ab
c)
r
ab
4
4
b)
r
2 | ab |
d)
r
| ab |
4
4
Solution :Tangent
g
at (a,
( , b)) to the circle x2 + y2 = r2 is
x
y
2
ax + by = r that is same as
+
=1
( a) ( b )
r
This meets x-axis at A
2
r
2
⎞
⎛r
⎜ , 0⎟
⎠
⎝a
and y-axis at B ⎛⎜ 0, r ⎞⎟
⎝ b⎠
1
∴ Area of ∆OAB = (OA)
2
2
2
Ans:- (b)
( )
(OB) =
1 r
2 | ab |
4