MATHEMATICS
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 =
1) 11
2) -13
3) 24
4) 100
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),
1 + 4 + 4 + 4 + c = 0
⇒ c= -13
Ans: (2)
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
1) x 2 + y 2 - 6x + 2y – 28 = 0
2) x 2 + y 2 – 6x + 2y + 28 = 0
3) x 2 + y 2 + 4x – 2y + 24 = 0
4) x 2 + y 2 + 2x – 2y - 12=0
Solution:
B
AB = 6 , OC perpendicular distance Afrom (3,
-1) to
C
2x – 5y + 18 = 0
=
6 + 5 + 18
= 29
O (3, -1)
4 + 25
2
2
r = OB = OC + BC = 29 + 9 = 38
∴ (x – 3)2 + (y + 1)2 = 38
⇒ x2 + y2 – 6x + 2y – 28 = 0
Ans: (1)
Q3.
The four distinct points, (0, 0), (2, 0), (0, -2) and
(k, –2) are concyclic if k =
1) 2
2) -2
3) 0
4) 1
Solution:
(2, 0) and (0, -2) are ends of the diameter
∴ Eqn is x2 + y2 – ax – by = 0
⇒ x2 + y2 – 2x + 2y = 0
Substituting (k, -2) , k2 + 4 – 2k – 4 = 0
k2 – 2k = 0, k =2 (k ≠ 0)
Ans: (1)
Q4.
The length of the intercept made by the
circle x2 + y2 + 10x – 12y – 13 = 0 on the
y – axis is
1) 1
3) 3
2) 2
4) 14
Solution:
Length = 2 f 2 - c = 2 36 + 13 = 14
Ans: (4)
Q5.
The length of the chord x = 3y + 13 of
the circle x2 + y2 – 4x + 4y + 3 = 0 is
1) 2 5
3) 3 5
2) 5 2
4) 10
Solution:
A
C
B
O (2,-2)
OB= radius of the circle =
4+4-3 = 5
OC = perpendicular distance from (2, -2) to
x - 3y – 13 = 0
= 2+6-13 = 5
1+9
10
2
2
A B = 2B C = 2 O B - O C = 2 5 - 2 5 = 1 0 =
10
10
Ans: (4)
10
Q6.
The area of the circle whose equations
are given by x = 3 + 2 cos θ and y = 1 + 2 sinθ
is
1) 4π
2) 6π
3) 9π
4) 8π
Solution:
Given x - 3 = 2cos θ and y – 1 = 2 sin θ.
⇒ (x – 3)2 + (y – 1)2 = 4 ⇒ radius = 2
∴ Area = 4π
∴ Ans: (1)
Q7.
A circle having area 9π sq. units touches both
the coordinate axes in the third quadrant, then
its equation is
1) x2 + y2 – 6x – 6y – 9 =0
2) x2 + y2 + 6x + 6y + 9 =0
3) x2 + y2 – 6x + 6y + 9 = 0
4) 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)2 + (y – k)2 = r2
⇒ (x + 3)2 + (y + 3)2 = 9
⇒ x2 + y2 + 6x + 6y + 9 = 0
3
•
(-3, -3)
Ans: (2)
3
Q8.
If x + y + k = 0 is a tangent to the circle
x2 + y2 – 2x – 4y + 3 = 0 then k =
1) ± 20
2) -1, -5
3) ± 2
4) 4
Solution:-
(1, 2)
x+y+k=0
radius =
1+ 4 - 3 = 2
Perpendicular distance from (1, 2) to x + y + k = 0} = radius
1+ 2 + k =
2
Ans(2)
2 ⇒ K + 3 = ±2 ⇒ k = -1, -5
Q9.
The radius of any circle touching the
lines 3x - 4y + 5 = 0 and 6x – 8y – 9 =0 is
1) 1
3) 20
19
2) 23
15
4) 19
20
Solution:
3x – 4y + 5 = 0 …………..1
6x – 8y – 9 = 0 ………………2
Multiply 1 by 2
6x – 8y + 10 = 0
Given lines are parallel lines
∴ 2r = distance between the lines
2r=
c - d = 10 + 9 = 19 ⇒ r = 19
10
20
a2 + b2
36 + 64
Ans: (4)
Q10.
The circles x2 + y2 – 10x + 16 = 0 and x2+ y2 = r2
intersect each other in distinct points if
1) r > 8
2) r < 2
3) 2 < r < 8
4) 2 ≤ r ≤ 8
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
1)
21
2
3) 2 21
2)
42
2
4)
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
1 + 1 + 10 = 42
4 4
2
Ans: (2)
∴r=
Q12.
The locus of the centre of a circle which
cuts the circles x2 + y2 + 4x – 6y + 9 = 0 and
x2 + y2 – 4x + 6y + 4 = 0 orthogonally is
1)
2)
3)
4)
12x + 8y + 5 = 0
8x + 12y + 5 = 0
8x – 12y + 5 = 0
12x – 8y – 5 = 0
Solution:
The locus is the radical axis of the circles.
∴ Radical axis is 8x – 12y + 5 = 0
Ans: (3)
Q13.
If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along
diameter of a circle of circumference 10π then the
equation of the circle is
1)
2)
3)
4)
x2
x2
x2
x2
+
+
+
+
y2
y2
y2
y2
– 2x + 2y – 23 = 0
– 2x – 2y – 23 = 0
+ 2x + 2y – 23 = 0
+ 2x – 2y – 23 = 0
Solution:
Solving 2x + 3y = -1
3x – y = 4
x3
-------------------2x + 3y = -1
9x – 3y = 12
-------------------11x = 11 ⇒ x = 1
∴ y = -1
∴ centre = (1, -1)
Also c = 2πr = 10π ⇒ r = 5
∴ Equation is (x – 1)2 + (y + 1)2 = 52
⇒ x2 + y2 – 2x + 2y – 23 = 0
∴ Ans: (1)
Q14.
If g2 + f2 = c then the equation x2 + y2 + 2gx + 2fy + c = 0
will represent
1)
2)
3)
4)
a
a
a
a
circle
circle
circle
circle
of
of
of
of
radius g
radius f
diameter
radius 0
c
Solution:
g2 + f2 = c ⇒ g2 + f2 – c = 0 ⇒ radius = 0
∴ Ans: (4)
Q15
The area of the circle 4x2 + 4y2 - 8x + 16y + k = 0 is
9π sq. units, then the value of k is
1) 4
3) -16
2) 16
4) ± 16
Solution:
k
÷ by 4, x + y – 2x + 4y +
=0
4
2
2
k
∴ radius = g + f − c = 1 + 4 −
4
But area = 9π ⇒ πr2 = 9π ⇒ r2 = 9 ⇒ r = 3
k
5
−
∴
=3
4
k
Squaring, 5 =9
4
k
i.e - = 4 ⇒ k = -16
4
2
Ans: (3)
2
Q16
The slope of the line perpendicular to the radical axis of
the circles x2 + y2 – 3x – 4y + 5 = 0 and
3x2 + 3y2 – 7x + 8y + 11 = 0 is
1) –3
3) 2
2) 10
3
4)
2
Solution:
The circles are x2 + y2 – 3x – 4y + 5 = 0 -------- (1)
7x
8y
11
x2 + y2 +
+
= 0 ---- (2)
3
3
3
Radical axis of (1) and (2) is (1) – (2)
7x
8y
11
–3x +
- 4y +5=0
3
3
3
⇒ –9x + 7x – 12y – 8y + 15 – 11 = 0
⇒ –2x – 20y + 4 = 0
( −2 )
1
Slope of radical axis = –
=–
− 10
10
∴ Slope of the line perpendicular to radical axis = 10
Ans: (2)
Q17
The area of the circle having centre at (3, 4) and
touching the line 5x + 12y – 11 = 0 is
1) 4π sq. units
3) 16π sq. units
2) 25π sq. units
4) 12π sq. units
Solution:
radius = perpendicular distance from (3, 4) to
5x + 12y – 11 = 0
(3, 4)
15 + 48 − 11
52
=
=
=4
r
13
25 + 144
∴ Area = πr2 = 16π sq. units
5x + 12y – 11 = 0
Ans: (3)
Q18.
A (4, 1) is a point on the circle x2 + y2 – 2x + 6y – 15 = 0.
Then the equation of tangent to the circle which
is parallel to the tangent at A is
1) 3x + y = 0
3) 3x – 10y – 34 = 0
2) 3x + 4y – 16 = 0
4) 3x + 4y + 34 = 0
A (4, 1)
Solution:
(1, -3) = midpoint of AB
4+ x 1+y
(1, -3) = 
,

2 
 2
⇒ x = –2, y = –7
∴ B = (-2, -7)
Equation of tangent at B is
xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0
x (-2) + y (-7) – 1 (x – 2) + 3 (y – 7) – 15 = 0
-2x – 7y – x + 2 + 3y – 21 – 15 = 0
-3x – 4y – 34 = 0
3x + 4y + 34 = 0
Ans: (4)
(1, -3)
B (x, y)
Q19.
The area of the largest square inscribed
in the circle x2 + y2 – 6x – 8y = 0 is
1) 100 sq units
3) 10 sq units
2) 25 sq units
4) 50 sq units
Solution:
g = -3., f = -4, c = 0
Radius =
5
•
a
a
9+16 = 5
∴ a2 + a2 = 102 ⇒ 2a2 = 100
⇒ a2 = 50
∴ Area = 50 sq units
∴ Ans: (4)
Q20.
If 2y + x + 3 = 0 is a tangent to 5x2 + 5y2 = k,
then the value of ‘K’ is
1) 4
2) 9
3) 16
4) 25
Solution:
5x2 + 5y2 = k
÷ by 5, x2+ y2 = k ……… (1)
5
y = -1 x -3
------------ (2)
2
2
(2) is a tangent to (1). Then c2= a2 (m2+1)
9 = k  1 +1 ⇒ 9 = k.5 ⇒ k = 9
5  4 
4
4
5.4
⇒k=9
Ans: (2)
Q21.
If the circles x2 + y2 + 4x – 4y - 6 = 0 and
x2 + y2 +kx + 2y + 8 = 0 cut orthogonally
then k =
1) 3
2) -3
3) 10
4) -10
Solution:
Condition is 2g1 g2 + 2f1f2 = c1+c2


k
i.e 2 (2 )   + 2 ( -2 ) .1= -6 + 8
 2 
2k – 4 = 2
∴ Ans: (1)
⇒ 2k = 6 ⇒ k = 3
Q22.
The radical centre of the circles x2 + y2 = 5,
‘x2 + y2- 3x +1 = 0 and x2 + y2 +2y – 1 =0 is
1) (-2, 2)
2) (2, 2)
3) (2, -2)
4) (1, 2)
Solution:
Radical axis of x2+ y2– 5 = 0 and 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
∴ Radical center = (2, -2)
Ans: (3)
Q23.
The number of common tangents to the
circles x2 + y2 + 2x + 8y – 23 = 0 and
x2 + y2 - 4x – 10y + 19 = 0 are
1) 1
2) 2
3) 3
4) 4
Solution:
C1 = (-1, -4), C2 = (2, 5),
r1 =
40 = 2 10 , r2 =
C1C 2 =
9+81 =
90 =3
10
10
= r1 + r 2
∴ Circles touch each other externally.
⇒ There are ‘3’ common tangents
∴ Ans: (3)
Q24.
The points from which the tangents to
the circles x2 + y2 – 8x + 40 = 0,
5x2 + 5y2 – 25x + 80 = 0 and
x2 + y2 – 8x + 16y + 160 = 0 are equal in
length is
 15 
1)  8,

2 

15 

3)  8,
2 

15 

2)  -8,

2 



-15

4)  -8,
2 

Q25.
Two circles of equal radius r cut orthogonally.
If their centres are (2, 3) and (5, 6) then r =
1) 1
2) 2
3) 3
4) 4
Solution:
C1 = (2, 3)
C2 = (5, 6)
∴ Eqn x2 + y2 – 4x – 6y + 13 – r2 = 0 …… (1)
x2 + y2 – 10x – 12y + 41 – r2 = 0 … (2)
(1) cuts (2) orthogonally
∴2g1 g2 + 2f1f2 = c1+ c2
OR
2(2) 5 + 2(3) 6 = 13 –r2 + 61 – r2
r
(2, 3)
r
•
• (5, 6)
20 + 36 = 74 – 2r2
2
2r = 74 – 56 = 18
⇒r =9 ⇒ r=3
2
Ans: (3)
2
2r =
2
2r =
(
(
2
2
(2 - 5) +(3 - 6)
9+9
)
2
= 18
)
2
Q26.
The number of tangents which can
be drawn from the point (1, 2) to
the circle x2 + y2 – 2x – 4y + 4 = 0 are
1) 1
2) 2
3) 3
4) 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: (4)
Q27.
The equation of the tangent to the circle
x2 + y2 = 9 which are parallel to the line
2x + y – 3 = 0 is
1) y = 2x ± 3 5
2) y = -2x ± 3 5
3) y = -2x
4) X + 2y = 0
Solution:
Let the equation of the tangent be y = mx+c
Since the tangent is parallel to the line
2x + y – 3 = 0, the slope of the tangent = -2
∴ m = -2 and radius is
r=3
But c2 = r2(m2 + 1)
⇒ c2 = 32[(-2)2 + 1] = 45
⇒c=±
45 = ± 3 5
∴ equation of tangent is y = mx + c
= -2x ± 3 5
Ans: (2)
Q28.
The length of the tangent from (3, -4) to the
circle 2x2 + 2y2 – 7x – 9y – 13 = 0 is
1)
3)
26
2
2) 26
4) 6
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 ) − (3) − ( − 4 ) −
2
2
2
=
21
13
9 + 16 −
+ 18 −
=
2
2
=
26
2
Ans: (1)
2
43 − 17
Q29.
The locus of the centre of a circle of radius
2 which rolls on the outside of the circle
x2 + y2 + 3x – 6y – 9 = 0 is
1) x2 + y2 + 3x – 6y + 5 = 0
2) x2 + y2 + 3x – 6y - 31 = 0
29
3) x + y + 3x – 6y +
=0
4
2
2
4) x2 + y2 + 3x + 6y + 29 = 0
Solution:
 3

Centre of given circle =  − , 3  = C
A
 2

B
9
81
9
C
+9+9 =
r=
=
= BC
4
4
2
AB = 2.
The locus of centre of outside circle is also a
circle concentric with given circle
9 13
=
∴ Its radius = AC = BA + BC = 2 +
2
2

 13 
 3 
2
Its equation is  x −  −   + (y – 3) = 

 2 
 2 

x2 + y2 + 3x – 6y – 31 = 0
Ans: (2)
2
2
Q30.
The area of the triangle formed by the
tangent at the point (a, b) to the circle
x2 + y2 = r2 and the coordinate axis is
r
1)
2ab
r
2)
2 | ab |
r
3)
ab
r
4)
| ab |
4
4
4
4
Solution:
Tangent at (a, b) to the circle x2 + y2 = r2 is ax
x
y
that is same as
+
=1
r
r
a
b
r

This meets x-axis at A  , 0  and y-axis at B
a

+ by = r2
( ) ( )
2
2
2
1
1 r
∴ Area of ∆OAB B=
(OA) (OB) =
2
2 | ab |
4
•(a, b)
Ans: (2)
O
A
r 

0
,


b


2
Q31
The locus of centres of family of circles passing through
the origin and cutting the circle x2 + y2 + 4x – 6y – 13 = 0
orthogoanlly is
1) 4x + 6y + 13 = 0
3) 4x + 6y – 13 = 0
2) 4x – 6y + 13 = 0
4) 4x – 6y – 13 = 0
Solution:
Equation is x2 + y2 + 2gx + 2fy = 0 --------------- (1)
(1) cuts x2 + y2 + 4x – 6y – 13 = 0 orthogonally
2 g1g2 + 2f1f2 = c1 + c2
2 g (2) + 2 f (-3) = 0 – 13
4 g – 6f + 13 = 0
Locus of (-g, -f) is
4 (-x) – 6 (-y) + 13 = 0
–4 x + 6y + 13 = 0
i.e. 4x – 6y – 13 = 0
Ans: (4)
Q32.
The centres of a set of circles each of radius 3 lie on the
circle x2 + y2 = 25. The locus of any point in the set is
1) 4 ≤ x2 + y2 ≤ 64
3) x2 + y2 ≥ 25
2) x2 + y2 ≤ 25
4) 3 ≤ x2 + y2 ≤ 9
Solution:
Let (h, k) be the centre of the circle.
Then equation is (x – h)2 + (y – k)2 = r2 = 9
The centre of the above circle lies on x2 + y2 = 25
∴ h2 + k2 = 25
∴ 5 – 3 ≤ distance between the centres of the two circles
≤5+3
2 ≤ h2 + k 2 ≤ 8
i.e. 4 ≤ h2 + k2 ≤ 64
∴ locus of (h, k) is 4 ≤ x2 + y2 ≤ 64
Ans: (1)
Q33.
If an equilateral triangle is inscribed in the circle x2 + y2 = 4
then the length of its side is
1)
3)
2
3
2) 2 2
4) 2 3
Solution:
O
OA = 2
AB
AB
cos 30 =
=
OA
2
AB
3
i.e.
=
⇒ AB =
2
2
∴side AC = 2 3
2
o
Ans: (4)
A
3
30o
B
C
Q34.
The line 4x + 3y – 25 = 0 touches the circle
x2 + y2 = 25 at the point
1) (1, 7)
 3 , 13 
3) 

3 

2) (4, 3)
4) (3, 4)
Solution:
 − a2 m a2 

point = 
,
 c
c 



− 4
 − 25 . 


 3  25 
,
=
= (4, 3)
25
25 



3
3


Ans: (2)
4x + 3y = 25
3y = -4x + 25
 25 
 4
x+



 3
 3 
y=  −