Circles: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Circle  
 
(January 2026) 
 
Q1: Let the circle x² + y² = 4 intersect x-axis at the points A(a, 0), a > 0 and B(b, 0). 
Let P(2 cos a, 2 sin a), 0 < a < p/2 and Q(2 cos ß, 2 sin ß) be two points such that 
(a - ß) = p/2. Then the point of intersection of AQ and BP lies on: 
A: x² + y² - 4x - 4 = 0 
B: x² + y² - 4x - 4y = 0 
C: x² + y² - 4x - 4y - 4 = 0 
D: x² + y² - 4y - 4 = 0 
Answer: D 
Explanation: 
 
The circle is x² + y² = 2² (radius r = 2). 
Intersection with x-axis: Setting y = 0 gives x² = 4 ? x = ±2. 
Since a > 0, A = (2, 0). 
Therefore, B = (-2, 0). 
Points P and Q : 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Circle  
 
(January 2026) 
 
Q1: Let the circle x² + y² = 4 intersect x-axis at the points A(a, 0), a > 0 and B(b, 0). 
Let P(2 cos a, 2 sin a), 0 < a < p/2 and Q(2 cos ß, 2 sin ß) be two points such that 
(a - ß) = p/2. Then the point of intersection of AQ and BP lies on: 
A: x² + y² - 4x - 4 = 0 
B: x² + y² - 4x - 4y = 0 
C: x² + y² - 4x - 4y - 4 = 0 
D: x² + y² - 4y - 4 = 0 
Answer: D 
Explanation: 
 
The circle is x² + y² = 2² (radius r = 2). 
Intersection with x-axis: Setting y = 0 gives x² = 4 ? x = ±2. 
Since a > 0, A = (2, 0). 
Therefore, B = (-2, 0). 
Points P and Q : 
P = (2 cos a, 2 sin a) 
Q = (2 cos ß, 2 sin ß) 
Let point of intersection of AQ & BP be R(h, k). 
So, we need to find locus of points of intersection of AQ & BP. 
Slope of BP = slope of BR 
 
Given, 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Circle  
 
(January 2026) 
 
Q1: Let the circle x² + y² = 4 intersect x-axis at the points A(a, 0), a > 0 and B(b, 0). 
Let P(2 cos a, 2 sin a), 0 < a < p/2 and Q(2 cos ß, 2 sin ß) be two points such that 
(a - ß) = p/2. Then the point of intersection of AQ and BP lies on: 
A: x² + y² - 4x - 4 = 0 
B: x² + y² - 4x - 4y = 0 
C: x² + y² - 4x - 4y - 4 = 0 
D: x² + y² - 4y - 4 = 0 
Answer: D 
Explanation: 
 
The circle is x² + y² = 2² (radius r = 2). 
Intersection with x-axis: Setting y = 0 gives x² = 4 ? x = ±2. 
Since a > 0, A = (2, 0). 
Therefore, B = (-2, 0). 
Points P and Q : 
P = (2 cos a, 2 sin a) 
Q = (2 cos ß, 2 sin ß) 
Let point of intersection of AQ & BP be R(h, k). 
So, we need to find locus of points of intersection of AQ & BP. 
Slope of BP = slope of BR 
 
Given, 
 
 
Q2: Let y = x be the equation of a chord of the circle C 1 (in the closed half-plane x 
= 0) of diameter 10 passing through the origin. Let C 2 be another circle described 
on the given chord as its diameter. If the equation of the chord of the circle C 2, 
which passes through the point (2, 3) and is farthest from the center of C 2, is x + 
ay + b = 0, then a - b is equal to 
A: -6 
B: 10 
C: 6 
D: -2 
Answer: D 
Explanation: 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Circle  
 
(January 2026) 
 
Q1: Let the circle x² + y² = 4 intersect x-axis at the points A(a, 0), a > 0 and B(b, 0). 
Let P(2 cos a, 2 sin a), 0 < a < p/2 and Q(2 cos ß, 2 sin ß) be two points such that 
(a - ß) = p/2. Then the point of intersection of AQ and BP lies on: 
A: x² + y² - 4x - 4 = 0 
B: x² + y² - 4x - 4y = 0 
C: x² + y² - 4x - 4y - 4 = 0 
D: x² + y² - 4y - 4 = 0 
Answer: D 
Explanation: 
 
The circle is x² + y² = 2² (radius r = 2). 
Intersection with x-axis: Setting y = 0 gives x² = 4 ? x = ±2. 
Since a > 0, A = (2, 0). 
Therefore, B = (-2, 0). 
Points P and Q : 
P = (2 cos a, 2 sin a) 
Q = (2 cos ß, 2 sin ß) 
Let point of intersection of AQ & BP be R(h, k). 
So, we need to find locus of points of intersection of AQ & BP. 
Slope of BP = slope of BR 
 
Given, 
 
 
Q2: Let y = x be the equation of a chord of the circle C 1 (in the closed half-plane x 
= 0) of diameter 10 passing through the origin. Let C 2 be another circle described 
on the given chord as its diameter. If the equation of the chord of the circle C 2, 
which passes through the point (2, 3) and is farthest from the center of C 2, is x + 
ay + b = 0, then a - b is equal to 
A: -6 
B: 10 
C: 6 
D: -2 
Answer: D 
Explanation: 
 
To find the value of a - b, 
Determine the equation of circle C 1 
The circle C 1 has a diameter of 10, so its radius is R = 5. It passes through the origin (0, 
0) and is contained in the closed half-plane x = 0. 
Let the center of C 1 be (h, k). Since it passes through (0, 0), we have h² + k² = R² = 25. 
For the circle to lie entirely in the region x = 0 while passing through (0, 0), the distance 
from the center to the y-axis must be equal to the radius, and the center must have a 
non-negative x-coordinate. 
Thus, h = R = 5. 
Substituting h = 5 into h² + k² = 25 gives 25 + k² = 25, which implies k = 0. 
The equation of C 1 is: 
(x - 5)² + y² = 25 
Find the center of circle C 2 
The line y = x is a chord of C 1. We find the intersection points of y = x and C 1 : 
(x - 5)² + x² = 25 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Circle  
 
(January 2026) 
 
Q1: Let the circle x² + y² = 4 intersect x-axis at the points A(a, 0), a > 0 and B(b, 0). 
Let P(2 cos a, 2 sin a), 0 < a < p/2 and Q(2 cos ß, 2 sin ß) be two points such that 
(a - ß) = p/2. Then the point of intersection of AQ and BP lies on: 
A: x² + y² - 4x - 4 = 0 
B: x² + y² - 4x - 4y = 0 
C: x² + y² - 4x - 4y - 4 = 0 
D: x² + y² - 4y - 4 = 0 
Answer: D 
Explanation: 
 
The circle is x² + y² = 2² (radius r = 2). 
Intersection with x-axis: Setting y = 0 gives x² = 4 ? x = ±2. 
Since a > 0, A = (2, 0). 
Therefore, B = (-2, 0). 
Points P and Q : 
P = (2 cos a, 2 sin a) 
Q = (2 cos ß, 2 sin ß) 
Let point of intersection of AQ & BP be R(h, k). 
So, we need to find locus of points of intersection of AQ & BP. 
Slope of BP = slope of BR 
 
Given, 
 
 
Q2: Let y = x be the equation of a chord of the circle C 1 (in the closed half-plane x 
= 0) of diameter 10 passing through the origin. Let C 2 be another circle described 
on the given chord as its diameter. If the equation of the chord of the circle C 2, 
which passes through the point (2, 3) and is farthest from the center of C 2, is x + 
ay + b = 0, then a - b is equal to 
A: -6 
B: 10 
C: 6 
D: -2 
Answer: D 
Explanation: 
 
To find the value of a - b, 
Determine the equation of circle C 1 
The circle C 1 has a diameter of 10, so its radius is R = 5. It passes through the origin (0, 
0) and is contained in the closed half-plane x = 0. 
Let the center of C 1 be (h, k). Since it passes through (0, 0), we have h² + k² = R² = 25. 
For the circle to lie entirely in the region x = 0 while passing through (0, 0), the distance 
from the center to the y-axis must be equal to the radius, and the center must have a 
non-negative x-coordinate. 
Thus, h = R = 5. 
Substituting h = 5 into h² + k² = 25 gives 25 + k² = 25, which implies k = 0. 
The equation of C 1 is: 
(x - 5)² + y² = 25 
Find the center of circle C 2 
The line y = x is a chord of C 1. We find the intersection points of y = x and C 1 : 
(x - 5)² + x² = 25 
? x² - 10x + 25 + x² = 25 
? 2x² - 10x = 0 
? 2x(x - 5) = 0 
The intersection points are (0, 0) and (5, 5). These are the endpoints of the diameter of 
circle C2. 
The center M of C 2 is the midpoint of the chord: 
M = ((0+5)/2, (0+5)/2) = (2.5, 2.5) 
Determine the equation of the chord of C 2 
We are looking for the equation of a chord of C 2 passing through P(2, 3) that is farthest 
from the center M(2.5, 2.5). 
In any circle, the chord passing through a point P that is farthest from the center is the 
one perpendicular to the radius (or segment) MP. 
The slope of the segment MP is: 
 
The slope m of the required chord is the negative reciprocal of m_MP : 
m = -1/-1 = 1 
Using the point-slope form for the line through P(2, 3) with slope m = 1 : 
y - 3 = 1(x - 2) 
? y - 3 = x - 2  
? x - y + 1 = 0 
Calculate a - b 
The equation is given in the form x + ay + b = 0. Comparing this to x - y + 1 = 0, we 
identify: 
a = -1  
b = 1 
Finally, we calculate the required value: 
a - b = -1 - 1 = -2 
 
Q3: Let a circle of radius 4 pass through the origin O, the points A(-v3a, 0) and 
B(0, -v2b), where a and b are real parameters and ab ? 0. Then the locus of the 
centroid of ?OAB is a circle of radius 
A: 7/3 
B: 11/3 
C: 5/3 
D: 8/3 
Answer: D