JEE ‘Legends’ (MCSIR) Circle

Q. Let the tangents drawn from the origin to the circle, x 2 + y2 – 8x – 4y + 16 = 0 touch it at the points
A and B. The (AB)2 is equal to: [Jee main 2020 (07-01-2020-shift-2)]
56 52 64 32
(A) (B) (C) (D)
5 5 5 5
Ans. (C)
Sol. L  S1  16  4

R  16  4  16  2

2LR 2 42 16
Length of chord of contact = L2  R 2  16  4  20

64
Square of length of chord of contact 
5

Q. If a line, y = mx + c is a tangent to the circle, (x – 3)2 + y2 = 1 and it is perpendicular to a line L1, where
 1 1 
L1 is the tangent to the circle, x2 + y2 = 1 at the point  ,  ; then :
 2 2
[Jee main 2020 (08-01-2020-shift-2)]
(A) c2 – 7c + 6 = 0 (B) c2 + 7c + 6 = 0 (C) c2 – 6c + 7 = 0 (D) c2+ 6c + 7 = 0
Ans. (D)
Sol. (x – 3)2 + y2 = 1, tangent is y =mx + c
 1 1 
for circle x2 + y2 = 1 tangnet at p  , 
 2 2
from T = 0, will be
1 1
x y 1  0
2 2
x+y– 2=0 ......L1
 x – y + l = 0  This is tangent to the circle (x – 3)2 + y2 = 1
r
x + y – 2 = 0 


apply r = p
30
1=  |l + 3| = 2
2
 = 3  2
(+ 3)2 = 2
2 + 9 + 6 – 2 = 0
c2 + 6c + 7 = 0

Q. A circle touches the y-axis at the point (0,4) and passes through the point (2,0). Which of the following
lines is not a tangent to this circle ? [Jee main 2020 (09-01-2020-shift-1)]
(A) 3x – 4y – 24 = 0 (B) 4x – 3y + 17 = 0 (C) 3x + 4y – 6 = 0 (D) 4x +3y – 8 = 0

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 1
JEE ‘Legends’ (MCSIR) Circle
Ans. (D)
x2   y  4   x  0
2
Sol.
Passes through (2, 0)
4  16  2  0
  10

x2  y2  8x  16  10x  0

C 5, 4  , r  25  16  16  5

Distance of (5,4) from line 4x + 3y – 8 = 0 ¹ radius

Option (4) is correct answer.

Q. If the curves, x2 – 6x + y2 + 8 = 0 and x2 – 8 y + y2 + 16 – k = 0, ( k > 0) touch each other at a point,

then the largest value of k is _______________. [Jee main 2020 (09-01-2020-shift-2)]
Ans. (36)
Sol. Two circle touches each other if C1C2 = r1  r2
Distance between C2(3,0) and C1(0,4) is either k  1 or k  1 (C C = 5)
1 2

 k  1  5 or k  1 = 5  k = 16 or k = 36

Q. The number of integral values of k for which the line, 3x+4y=k intersects the circle, x2 + y2 – 2x – 4y + 4 = 0
at two distinct points is................ [Jee main 2020 (02-09-2020-shift-1)]
Ans. (9)
Sol. c : (1,2) & r = 1
|cp| < r
3.1  4.2  k
1
5
|11– k| < 5
– 5< k –11 < 5
6 < k < 16
k = 7, 8, 9, ......., 15  total 9 value of k

Q. The set of all possible values of  in the interval (0,  ) for which the points (1,2) and  sin ,cos   lie
on the same side of the line x + y = 1 is : [Jee main 2020 (02-09-2020-shift-2)]
     3    3 
(A)  0,  (B)  0,  (C)  0,  (D)  , 
 4  2  4  4 4 

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 2
JEE ‘Legends’ (MCSIR) Circle
Ans. (B)
Sol. (sinq, cosq) lie on x2  y2  1 (1,2)

Shaded points satisfy   (0,  / 2)

L : x+y=1

Q. The diameter of the circle, whose centre lies on the line x + y = 2 in the first quadrant and which touches
both the lines x = 3 and y = 2, is .......... [Jee main 2020 (03-09-2020-shift-1)]
Ans. (2)
Sol. p=r
for y = 2
22 y=2
r=   r
1
r
for x = 3
3
r=  3
1 x=3
|| = |a – 3|
3
 2 + 2 - 6 + 9   =
2
2 = 3 = 2r

Q. The circle passing through the intersection of the circles, x2 + y2 – 6x = 0 and x2 + y2 – 4y = 0, having its
centre on the line, 2x – 3y + 12=0, also passes through the point :
[Jee main 2020 (04-09-2020-shift-2)]
(A) (–1,3) (B) (1,–3) (C) (–3,6) (D) (–3,1)
Ans. (D)
Sol. Sol. (3)
S1 + (S1 – S2) = 0
x2 + y2 – 6x + (4y – 6x) = 0
x2 + y2 – 6x(1 + ) + 4y = 0
Centre (3(1 +), – 2) put in 2x – 3y + 12 = 0
6 + 6+ 6+ 12 = 0
12= – 18
= – 3/2
Circle is x2 + y2 + 3x – 6y = 0
Check options

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 3
JEE ‘Legends’ (MCSIR) Circle

Q. Let PQ be a diameter of the circle x2 + y2 = 9. If  and  are the lengths of the perpendiculars from P
and Q on the straight line, x + y = 2 respectively, then the maximum value of αβ is _____
[Jee main 2020 (04-09-2020-shift-2)]
Ans. (7)

Q(acos ,asin )

Sol.
P
(–acos ,-asin )
x+y=2

3cos   3sin   2

2
3cos   3sin   2

2

 3cos   3sin  
2
4 9  9 sin 2  4 5  9 sin 2
       
2 2 2

95
 max  7
2

Q. If the length of the chord of the circle, x2 + y2 = r2(r > 0) along the line, y – 2x = 3 is r, then r2 is equal
to : [Jee main 2020 (05-09-2020-shift-2)]
24 9 12
(A) 12 (B) (C) (D)
5 5 5
Ans. (D)
Sol. AB = 2 r 2  9 / 5 =r

r2 3
M
r —9/5 =
2
5 A
4 r
3r2/4 = 9/5 (0,0)

12
r2 
5

2 2
Q. If one of the diameters of the circle x + y – 2x – 6y + 6 = 0 is a chord of another circle ‘C’, whose
center is at (2, 1), then its radius is ______ . [Jee main 2021 (24-02-2021-shift-1)]
Ans. (3)
2 2
Sol. x + y – 2x – 6y + 6 = 0
center (1, 3)

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 4
JEE ‘Legends’ (MCSIR) Circle

radius = 2
distance between (1, 3)and(2, 1) is 5
 ( 5)2  (2)2  r 2
r=3

Q. Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point
2
(– 5, 0). If the locus of the point P is a circle of radius r, then 4r is equal to ________.
[Jee main 2021 (24-02-2021-shift-2)]
Ans. (56)
Sol. Let point is (h, k)
So, (h – 5)2  k 2  3 (h  5)2  k 2
2 2
8x + 8y + 100 x + 200 = 0
25
x2  y2  x  25  0
2

(25)2
r2  – 25
42
(25)2
4r 2  – 100
4
2
4r = 156.25 – 100
2
4r = 56.25
2
After round of 4r = 56

Q. If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle
2 2
(x – 2) + (y – 3) = 25 at the point (5, 7) is A, then 24A is equal to_____.
[Jee main 2021 (24-02-2021-shift-2)]
Ans. (1225)
Sol. (Bonus)
73
y3  x  2
52
3y  9  4x  8

N :4x  3y  1  0
1
Put Put y  0  x  
4
 1 
  , 0  Not possible
 4 

* as positive x-axis is given in the question so question should be bonus.

Q. In the circle given below, let OA = 1 unit, OB = 13 unit and PQ  OB. Then, the area of the triangle
PQB (in square units) is : [Jee main 2021 (26-02-2021-shift-1)]

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 5
JEE ‘Legends’ (MCSIR) Circle

(A) 24 2 (B) 24 3 (C) 26 3 (D) 26 2

Ans. (B)
Sol. Let PA = AQ = 
OA · AB = AP · AQ
 1.12 =  · 
2 3
1
Area PQB   2  AB
2
1
  4 3  12
2
 24 3
2 2
Q. If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x + y = 1
is a circle of radius r, then r is equal to : [Jee main 2021 (26-02-2021-shift-2)]
1 1 1
(A) 1 (B) (C) (D)
2 3 4
Ans. (B)
cos   3
Sol. h
2
(cos, sin)
sin   2
k
2
2
 3 1 (3,2)
  h     k  1 
2

 2 4

1
r 
2

2 2
Q. Let A(1, 4) and B(1, –5) be two points. Let P be a point on the circle (x – 1) + (y – 1) = 1 such that
2 2
(PA) + (PB) have maximum value, then the points P, A and B lie on :
(A) a straight line (B) a hyperbola (C) an ellipse (D) a parabola
[Jee main 2021 (26-02-2021-shift-2)]
Ans. (A)
2 2
Sol. P be a point on (x – 1) + (y – 1) = 1
so P(1 + cos, 1 + sin)
A(1,4) B(1,–5)
2 2
(PA) + (PB)

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 6
JEE ‘Legends’ (MCSIR) Circle
2 2 2 2
= (cos) + (sin – 3) + (cso) + (sin + 6)
= 47 + 6sin
is maximum if sin = 1
 sin = 1, cos = 0
P(1,2) A(1,4) B(1,–5)
P, A, B are collinear points.

Q. Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes
2 2
through (3, –3) and (4, 2 2) , and given that a  2 2b  3 , then (a + b + ab) is equal to ______.
[Jee main 2021 (26-02-2021-shift-2)]
Ans. (9)
Sol. All normals of circle passes through centre
Radius = CA = CB
2 2
CA = CB
2 2
(a – 3) + (b + 3)

 
2
 (a  4)2  b  2 2


a 32 2b  3

a  2 2 b  3b  3 ...(1)
given that a  2 2 b  3 ...(2)
from (1) & (2)  a = 3, b = 0
2 2
a + b + ab = 9

Q. Let ABCD be a square of side of unit length. Let a circle C centered at A with unit radius is drawn.
1
Another circle C which touches C and the lines AD and AB are tangent to it, is also drawn. Let a
2 1
tangent line from the point C to the circle C meet the side AB at E. If the length of EB is   3 ,
2
where  are integers, then  is equal to_______. [Jee main 2021 (16-03-2021-shift-1)]
Ans. (1)
Sol. Here AO + OD = 1 or ( 2  1)r  1
 r  2 1
2 2 2
equation of circle (x – r) + (y – r) = r

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 7
JEE ‘Legends’ (MCSIR) Circle

C 1,1
2 − 1, 2 − 1 θ
O
P
45°
A B
E

OP 2 1
sin   
OC
 
2
2 2 2

2 1


2 2 2 

2 1 1
 
2  2 1  2

  30
CEB  75
EB
cot 75 
1
EB  2  3
  2 ;   1
  1

EB  2  3

2 2
Q. Let the lengths of intercepts on x-axis and y-axis made by the circle x + y + ax + 2ay + c = 0, (a < 0)
be 2 2 and 2 5 , respectively. Then the shortest distance from origin to a tangent to this circle which
is perpendicular to the line x + 2y = 0, is equal to : [Jee main 2021 (16-03-2021-shift-2)]
(A) 11 (B) 7 (C) 6 (D) 10
Ans. (C)
2 2
Sol. x + y + ax + 2ay + c = 0
a2
2 g2  c  2 c  2 2
4

a2
 c  2 ….(1)
4

2 f 2  c  2 a2  c  2 5
2
 a –c=5 ….(2)
(1) & (2)

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 8
JEE ‘Legends’ (MCSIR) Circle

3a 2
 3  a  2 (a  0)
4
 c = –1
2 2
Circle  x + y – 2x – 4y – 1 =0
2 2
 (x – 1) + (y – 2) = 6
1
Given x  2y  0  m  
2
m =2
tangent
Equation of tangent
 (y  2)  2(x  1)  6 1  4

 2x  y  30  0

 30
Perpendicular distance form (0. 0)   6
4 1

Q. The line 2x – y + 1 = 0 is a tangent to the circle at the point (2, 5) and the centre of the circle lies on
x – 2y = 4. Then, the radius of the circle is: [Jee main 2021 (17-03-2021-shift-1)]
(A) 3 5 (B) 5 3 (C) 5 4 (D) 4 5
Ans. (A)

Sol.

 h4 
5 2 
  (2)  1
 2  h 
 
h=8
Center (8, 2)
Radius  36  9  3 5

Q. Choose the incorrect statement about the two circles whose equations are given below :
2 2
x + y – 10x – 10y + 41 = 0 and
2 2
x + y – 16x – 10y + 80 = 0 [Jee main 2021 (17-03-2021-shift-1)]
(A) Distance between two centres is the average of radii of both the circles.
(B) Both circles centres lie inside region of one another.
(C) Both circles pass through the centre of each other.
(D) Circles have two intersection points.

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 9
JEE ‘Legends’ (MCSIR) Circle

Ans. (B)
Sol. r = 3, c (5, 5)
1 1
r = 3, c (8, 5)
2 2
C C = 3, r = 3, r = 3
1 2 1 2

Q. The minimum distance between any two points P and P while considering point P on one circle and
1 2 1
point P on the other circle for the given circles’ equations
2 22
x + y –10x – 10y + 41 = 0
2 2
x + y – 24x – 10y + 160 = 0 is ______ . [Jee main 2021 (17-03-2021-shift-1)]
Ans. (1)
Sol. Given C (5, 5), r = 3 and C (12, 5), r = 3
1 1 2 2
Now, C C > r + r
1 2 1 2
Thus, (P P ) =7–6=1
1 2 min

2 2
Q. Two tangents are drawn from a point P to the circle x + y – 2x – 4y + 4 = 0, such that the angle
 12   12 
between these tangents is tan 1   , where tan 1    (0, ). If the centre of the circle is denoted
 5  5
by C and these tangents touch the circle at points A and B, then the ratio of the areas of PAB and
CAB is : [Jee main 2021 (17-03-2021-shift-2)]
(A) 11 : 4 (B) 9 : 4 (C) 3 :1 (D) 2 : 1
Ans. (B)
12
Sol. tan  
5

PA  cot
2
1 1 
 area of PAB  (PA) sin   cot sin 
2 2

2 2 2 A
P
1  1  cos   1 
  sin 
2  1  cos   C
(1,2) 1
 5 
1 B
1 13   12   1  18  12  27
  
2 5   13  2 8 13 26
 1 
 13 

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 10
JEE ‘Legends’ (MCSIR) Circle

1 1  12  6
area of CAB  sin     
2 2  13  13
area of PAB 9
 
area of CAB 4

2 2
Q. Let the tangent to the circle x + y = 25 at the point R(3, 4) meet x-axis and y-axis at point P and Q,
respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre
2
of the triangle OPQ, then r is equal to [Jee main 2021 (17-03-2021-shift-2)]
529 125 625 585
(A) (B) (C) (D)
64 72 72 66
Ans. (C)
Sol. Tangent to circle 3x + 4y = 25

25
Q 0,
4

25 𝑎
b=
4 25
P ,0
3
O 0,0 25
c=
3

OP + OQ + OR = 25
 ax  bx 2  cx3 ay1  by2  cy3 
Incentre   1 ,
 a bc a  b  c 

 25 25 25 25 
  
Incentre   4 3 , 4 3 

 25 25 

 

 25 25 
 , 
 12 12 
2
 25  625 625
r  2 
2
  2 
 12  144 72
Q. Choose the correct statement about two circles whose equations are given below :
2 2
x + y – 10x – 10y + 41 = 0
2 2
x + y – 22x – 10y + 137 = 0 [Jee main 2021 (18-03-2021-shift-1)]
(A) circles have same centre (B) circles have no meeting point
(C) circles have only one meeting point (D) circles have two meeting points

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 11
JEE ‘Legends’ (MCSIR) Circle
Ans. (C)
2 2
Sol. x + y –10x– 10y + 41 = 0
A(5,5), R = 3
1
2 2
x + y – 22x – 10y + 137 = 0
B(11,5), R = 3
2
AB = 6 = R + R
1 2
Touch each other externally
circles have only one meeting point.

Q. For the four circles M, N, O and P, following four equations are given :
2 2
Circle M : x + y = 1
2 2
Circle N : x + y – 2x = 0
2 2
Circle O : x + y – 2x – 2y + 1 = 0
2 2
Circle P : x + y – 2y = 0 [Jee main 2021 (18-03-2021-shift-1)]
If the centre of circle M is joined with centre of the circle N, further centre of circle N is joined with
centre of the circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P
is joined with centre of circle M, then these lines form the sides of a :
(A) Rhombus (B) Square (C) Rectangle (D) Parallelogram
Ans. (B)
2 2
Sol. M:x +y =1 (0,0)
2 2
N : x + y – 2x = 0 (1,0)
2 2
O : x + y – 2x – 2y + 1 = 0 (1,1)
2 2
P : x + y – 2y = 0 (0,1)

2 2 2 2
Q. Let S : x + y = 9 and S : (x – 2) + y = 1. Then the locus of center of a variable circle S which
1 2
touches S internally and S externally always passes through the points :
1 2
[Jee main 2021 (18-03-2021-shift-2)]
1 5  3
(A) (0,  3) 
(B)  2 ,   (C)  2,   (D) (1, ± 2)
 2   2
Ans. (C)

Sol.

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 12
JEE ‘Legends’ (MCSIR) Circle

c c =r –r
1 2 1 2

 given circle are touching internally

Let a variable circle with centre P and radius r
 PA = r – r and PB = r + r
1 2
 PA + PB = r + r
1 2
 PA + PB = 4 ( > AB)
 Locus of P is an ellipse with foci at A(0, 0) and B(2, 0) and length of major axis is 2a = 4,
1
e
2
2 2 2
 centre is at (1. 0) and b = a (1 – e ) = 3

(x  1)2 y 2
E:  1
4 3

 3
which is satisfied by  2,  
2
 

Q. Let r1 and r2 be the radii of the largest and smallest circles, respectively, which pass through the point
(–4, 1) and having their centres on the circumference of the circle
r1
x2 + y2 + 2x + 4y – 4 = 0. If r  a  b 2 , then a + b is equal to :
2
[Jee main 2021 (20-07-2021-shift-2)]
(A) 7 (B) 11 (C) 5 (D) 3
Ans. (C)
Sol. x2 + y2 + 2x + 4y – 4 = 0
(x + 1)2 + (y + 2)2 = 33
General point on circumference (3 cos  – 1, 3 sin – 2)
As centre of circle is (–1, –2)

So, r  (3cos   3) 2  (3sin   3) 2

 3 cos 2   1  2 cos   sin 2   1  2sin 

 3 3  2(cos   sin )

r1  3 3  2 2

r2  3 3  2 2

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 13
JEE ‘Legends’ (MCSIR) Circle

r1 3 3  2 2
  3 2 2
r2 3 3  2 2

r1
On comparing with r  a  b 2
2

a+b=5

Q. Let the circle S: 36x2 + 36y2 – 108x + 120y + C = 0 be such that it neither intersects nor touches the co-
ordinates axes. If the point of intersection of the lines, x – 2y = 4 and 2x – y = 5 lies inside the circle S,
then: [Jee main 2021 (22-07-2021-shift-2)]
25 13
(A) C (B) 100 < C < 165 (C) 100 < C < 156 (D) 81 < C < 156
9 3
Ans. (C)
Sol. Intersection point of 2x – y = 5 and x – 2y = 4 is (2, –1)
So, (2, –1) lies inside the circle  S1 < 0
36(2)2 + 36 (–1)2 – 108 (2) + 120 (–1) + c < 0
c < 156 …….(i)
 circle 36x2 + 36y2 – 108x + 120y + c = 0 neither touches nor cuts the co-ordinate axis so
2
 –3  c
g2 –c<0   –  0  c  81 ….(ii)
 2  36
2
5 c
and f2 – c < 0    – < 0  c > 100 ….(iii)
 3  36
From (i), (ii) and (iii)
100 < c < 156

Q. Let [Jee main 2021 (27-07-2021-shift-1)]

A = {(x, y)  R × R | 2x + 2y – 2x – 2y = 1},
2 2

B = {(x, y)  R × R | 4x2 + 4y2 – 16y + 7 = 0} and

C = {(x, y)  R × R | x2 + y2 – 4x – 2y + 5  r2}.
Then the minimum value of |r| such that A  B  C is equal to :
3  10 2  10 3 2 5
(A) (B) (C) (D) 1  5
2 2 2
Y
Ans. (C)
Sol. Let centre of circle C is P B
r=3/2 A
Centre of circle B is Q (0,2) (2,1)
P
For A  B  C (1,1)
X
C

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 14
JEE ‘Legends’ (MCSIR) Circle
3
r  PQ +
2

3
r  2 1 
2 2

3 2 5 3 2 5
r  rmin =
2 2
Q. Let P and Q be two distinct points on a circle which has centre at C(2, 3) and which passes through
origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P, Q} is equal
to : [Jee main 2021 (27-07-2021-shift-1)]
(A) {(–1, 5), (5, 1)} (B) {(2 + 2 2 , 3 – 5 ), (2 – 2 2 , 3 + 5 )}
(C) {(2 + 2 2 , 3 + 5 ), (2 – 2 2 , 3 – 5 )} (D) {(4, 0), (0, 6)}
Ans. (A)
Sol. (x – 2)2 + (y – 3)2 = 13 P
Equation of line OC
C(2, 3)
3
y x
2
(0 0) Q
Line perpendicular to the above line and passing through (2, 3) is 3y + 2x = 13
Coordinates of P, Q 
 2  13 cos ,3  13 sin  
  3   2 
  2  13   ,3  13  
  13   13  
 (–1, 5) & (5, 1)

Q. Two tangents are drawn from the point P(–1, 1) to the circle x2 + y2 – 2x – 6y + 6 = 0. If these tangents
touch the circle at point A and B, and if D is a point on the circle such that length of the segments AB and
AD are equal, then the area of the triangle ABD is equal to: [Jee main 2021 (27-07-2021-shift-1)]
(A) 4 (B) 2 (C) 3( 2  1) (D) 3( 2  2)

Ans. (A)
Sol.
y

(1,3)
(–1,3) B D
2 2 2 2

P
(–1,1) A(1,1)
x
O

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 15
JEE ‘Legends’ (MCSIR) Circle

1
PAB =  4 2  4
2

Q. Consider a circle C which touches the y-axis at (0, 6) and cuts off intercept 6 5 on the x-axis. Then the
radius of the circle C is equal to : [Jee main 2021 (27-07-2021-shift-2)]
(A) 8 (B) 53 (C) 9 (D) 82
Ans. (C)
Sol. AD = 3 5
y
CA2 = CD2 + AD2
= 36 + 45 (0, 6) r C
P
CA2 = 81
D x
CA = 9 A
 r=9

Q. The locus of a point, which moves such that the sum of squares of its distance from the points (0,0),
(1,0), (0,1), (1,1) is 18 units, is a circle of diameter d. Then d2 is equal to ______.
[Jee main 2021 (26-08-2021-shift-1)]
Ans. (16)
Sol. Let P(x, y)
x 2  y 2   x  0    y  1   x  1   y  0    x  1   y  1  18
2 2 2 2 2 2

4x 2  4y 2  4x  4y  4  18
14
x 2  y2  x  y  0
4
1 1 1 1 14
Centre  ,  r   2
2 2 4 4 4
d  2r  2  2  4
So, d 2  16

Q. A circle C touches the line x = 2y at the point (2, 1) and intersects the circle C1 : x 2  y 2  2y  5  0 at
two points P and Q such that PQ is a diameter of C1 . Then the diameter of C is:
[Jee main 2021 (26-08-2021-shift-2)]
(A) 4 15 (B) 7 5 (C) 15 (D) 245
Ans. (D)
Sol. Family of circle touching line 2y = x at point (2, 1)
 x  2    y  1    x  2y   0
2 2
…… (1)
Common chord PQ is
 x  2    y  1    x  2y   x 2  y 2  2y  5  0
2 2

(Diameter of C1 passes through (0, –1))

4  4  2  1  2  5  0
Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 16
JEE ‘Legends’ (MCSIR) Circle
  7 put in (1)
 x  2    y  1  7  x  2y   0
2 2

x 2  y 2  11x  12y  5  0

121 121  124 245 245

r  36  5     diameter  245
4 4 4 2

Q. Let the equation x 2  y 2  px  11  p  y  5  0 represent circles of varying radius r   0,5 . Then
the number of elements in the set S  {q : q  p 2 and q is an integer} is ______.
[Jee main 2021 (27-08-2021-shift-1)]
Ans. (61)

p 2 1  p 
2
2p 2  2p  19
Sol. r  5 
4 4 4
Since, r   0, 5
So, 0  2p 2  2p  19  100

1  239 1  39   1  39 1  239 
 p ,   , 
 2 2   2 2 
So, number of integral values of p 2 is 61.

Q. Let Z be the set of all integers,

A  x, y   Z  Z;  x  2 2
 y2  4
B   x, y   Z  Z; x 2  y 2  4 and

C  x, y   Z  Z;  x  2  2
  y  2  4
2
 [Jee main 2021 (27-08-2021-shift-2)]
If the total number of relations from A  B to A  C is 2p , then the value of p is:
(A) 25 (B) 9 (C) 16 (D) 49
Ans. (A)
Sol. 
A   x, y   Z  Z;  x  2   y 2  4
2

B   x, y   Z  Z; x  y  4
2 2

C  x, y   Z  Z;  x  2  2
  y  2  4
2


Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 17
JEE ‘Legends’ (MCSIR) Circle

Similarly n  A  C   5
Relation from A  B to A  C  255  2 p  p  25

Q. Two circles each of radius 5 units touch each other at the point (1, 2). If the equation of their common
tangents is 4x + 3y = 10, and C1  ,   and C2  ,   , C1  C2 are their centres, then         
is equal to ______. [Jee main 2021 (27-08-2021-shift-2)]
Ans. (40)
4
Sol. Slope of the common tangent = 
3
4 3
If C1C2 makes an angle  with x-axis, then cos   and sin   .
5 5

So, the equation of in parametric form is

x 1 y  2
 ….(i)
4/5 3/5
Since, C1 and C2 are points on Eq. (i) at a distance of 5 units from P.
So, the coordinates of C1 and C2 are given by
x 1 y  2
  5
4/5 3/5

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 18
JEE ‘Legends’ (MCSIR) Circle

 x  1  4 and y  2  3.
Thus, the coordinates of C1 and C2 are (5, 5) and (–3, –1) respectively.
Hence, the equations of the two circles are
(x  5) 2  (y  5) 2  52
and (x  3) 2  (y  1) 2  52
C2  5,5  , C2  3, 1

          40

 x  1   y  1
2 2
Q. If the variable line 3x + 4y =  lies between the two circles  1 and

 x  9    y  1
2 2
 4 , without intercepting a chord on either circle, then the sum of all the integral
values of  is ______. [Jee main 2021 (31-08-2021-shift-1)]
Ans. (65)
Sol. Line lies between the two circle
(10 – ) (31 – ) < 0
10 <  < 31 …….(1)

As well as line can be tangent to the circle

P1  (Distance of (1, 1) from line)
P2  (Distance of (9, 1) from line)
P1  r1  7    10

    , 2  12,   …… (1)

P2  r2   31     10     , 21   41,  

1   2    3  12    21
10
Sum of integer value of   12  13   21   33  165
2

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 19
JEE ‘Legends’ (MCSIR) Circle

Q. Let B is the centre of the circle x 2  y 2  2x  4y  1  0 . Let the tangents at two points P and Q on the
 
circle intersect at the point A(3, 1). Then 8   Area APQ  is equal to
 Area BPQ 
[Jee main 2021 (31-08-2021-shift-2)]
Ans. (18)
Sol. AP  9  1  6  4  1
AP  3  AQ
r  1 4 1  2
3 Area APQ AR 3sin  9  Area APQ 
tan        8   18
2 Area BPQ RB 2cos  4  Area BPQ 

Get 10% Instant Discount On Unacademy Plus [Use Referral Code: MCSIR] 20