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All questions of Circle for JEE Exam

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 The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Poonam Reddy answered
 Radius of circle is given by -
r = √[(h-x1)² + (k-y1)²]
r = √[(2-3)² + (1+5)²]
r = √(-1² + 6²)
r = √(1 + 36)
r = √37
if centre (2,-]1) and radius=√26 are given,
(x-h)2+(y-k)2=r2
equation is (x-2)2 + (y-1)2 = (√37)2
x2 + 4 - 4x + y2 + 1 - 2y = 37
x2 + y2 - 4x - 2y - 32 = 0

The equation of the circle passing through (0, 0) and making intercepts 2 and 4 on the coordinate axes is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Geetika Shah answered
The circle intercept the co-ordinate axes at a and b. it means x - intercept at ( a, 0) and y-intercept at (0, b) .
Now, we observed that circle passes through points (0, 0) , (a, 0) and (0, b) .
we also know, General equation of circle is
x² + y² + 2gx + 2fy + C = 0
when point (0,0)
(0)² + (0)² + 2g(0) + 2f(0) + C = 0
0 + 0 + 0 + 0 + C = 0
C = 0 -------(1)
when point (a,0)
(a)² + (0)² + 2g(a) + 2f(0) + C = 0
a² + 2ag + C = 0
from equation (1)
a² + 2ag = 0
a(a + 2g) = 0
g = -a/2
when point ( 0, b)
(0)² + (b)² + 2g(0) + 2f(b) + C = 0
b² + 2fb + C = 0
f = -b/2
Now, equation of circle is
x² + y² + 2x(-a/2) + 2y(-b/2) + 0 = 0 { after putting values of g, f and C }
x² + y² - ax - by = 0
As we know that, a=2, b=4
x^2 + y^2 - 2x - 4y = 0

 The centre and radius of the circle x2 + y2 + 4x – 6y = 5 is:
  • a)
    (2, – 3), 2√2
  • b)
    (– 2, 3), 3√2
  • c)
    (– 2, 3), 2√2
  • d)
    (2, – 3), 3√2
Correct answer is option 'B'. Can you explain this answer?

Riya Banerjee answered
x2+y2+4x-6y=5
Circle Equation
(x-a)2+(y-b)2=r2 is the circle equation with a radius r, centered at (a,b)
Rewrite x2+y24x-6y=5 in the form of circle standard circle equation
(x-(-2))
2
+(y-3)2=(3
√2)2
Therefore the circle properties are:
(a,b) = (-2,3), r = 3√2

The locus of the mid-points of the chords of the circle x2 + y2 – 2x – 4y – 11 = 0 which subtend 60º at the centre is
  • a)
    x2 + y2 – 4x – 2y – 7 = 0
  • b)
    x2 + y2 + 4x + 2y – 7 = 0
  • c)
    x2 + y2 – 2x – 4y – 7 = 0
  • d)
    x2 + y2 + 2x + 4y + 7 = 0
Correct answer is option 'C'. Can you explain this answer?

Preeti Iyer answered
Let AB be the chord of the circle and P be the midpoint of AB.
It is known that perpendicular from the center bisects a chord.
Thus △ACP is a right-angled triangle.
Now AC=BC= radius.
The equation of the give circle can be written as
(x−1)2+(y−2)2=16
Hence, centre C=(1,2) and radius =r=4 units.
PC=ACsin60degree
= rsin60degree
= 4([2(3)½]/2
= 2(3)1/2 units
Therefore, PC=2(3)1/2
⇒ PC2=12
⇒ (x−1)2+(y−2)2=12
⇒ x2+y2−2x−4y+5=12
⇒ x2+y2−2x−4y−7=0

y = √3x + c1 & y = √3x + c2 are two parallel tangents of a circle of radius 2 units, then |c1 – c2| is equal to
  • a)
    8
  • b)
    4
  • c)
    2
  • d)
    1
Correct answer is option 'A'. Can you explain this answer?

Anmol Chauhan answered
For both lines to be parallel tangent the distance between both lines
should be equal to the diameter of the circle
⇒ 4 = |c1−c2|/(1+3)1/2
⇒∣c1−c2∣ = 8

The equation of the circle having the lines y2 – 2y + 4x – 2xy = 0 as its normals & passing through the point (2, 1) is
  • a)
    x2 + y2 – 2x – 4y + 3 = 0
  • b)
    x2 + y2 – 2x + 4y – 5 = 0
  • c)
    x2 + y2 + 2x + 4y – 13 = 0
  • d)
    None
Correct answer is option 'A'. Can you explain this answer?

Geetika Shah answered
The normal line to circle is →y² - 2 y + 4 x -2 xy=0
→ y(y-2) - 2x(y-2)=0
→ (y-2)(y-2x)=0
the two lines are , y=2 and 2 x -y =0
The point of intersection of normals are centre of circle.
→ Put , y=2 in 2 x -y=0, we get
→2 x -2=0
→2 x=2
→ x=1
So, the point of intersection of normals is (1,2) which is the center of circle.
Also, the circle passes through (2,1).
Radius of circle is given by distance formula = [(1-2)² + (2-1)²]½ 
=(1+1)½ =(2)½
The equation of circle having center (1,2) and radius √2 is
= (x-1)²+(y-2)²=[√2]²
→ (x-1)²+(y-2)²= 2
x²+y² -2x-4y+3 = 0

The equations of the tangents drawn from the point (0, 1) to the circle x2 + y2 - 2x + 4y = 0 are
  • a)
    2x - y + 1 = 0, x + 2y - 2 = 0
  • b)
    2x - y - 1 = 0, x + 2y - 2 = 0
  • c)
    2x - y + 1 = 0, x + 2y + 2 = 0
  • d)
    2x - y - 1 = 0, x + 2y + 2 = 0
Correct answer is option 'A'. Can you explain this answer?

Geetika Shah answered
Let equation of tangent with slope =m and point (0,1)
(y−1)=m(x−0)⇒y=mx+1
Intersection point
x2+(mx+1)2−2x+4(mx+1)=0
(1+m2)x2+(−2+6m)x+5=0
For y=mx+1 to be tangent, discriminant =0
(6m−2)2−4×5(1+m2)=0
36m2+4−24m−20m2+20=0
16m2−20m+24=0
⇒ 2m2−3m−2=0
(2m+1)(m−2)=0

The circle with (4, –1) and touching x-axis is
  • a)
    x2 + y2 - 8x + 2y + 16 = 0
  • b)
    x2 + y2 + 18x - 2y -16 = 0
  • c)
    x2 + y2 - 4x + y + 4 = 0
  • d)
    x2 + y2 + 14x - y + 4 = 0
Correct answer is option 'A'. Can you explain this answer?

Om Desai answered
The centre of given circle = (4, -1)
The circle is touching the X-axis.
Then the equation of the circle-
(x-h)2 + (y-k)2 = r2
Where (h, k) = (4, -1)
r = radius of circle
Since the circle is touching the X-axis. Then
r = k
r = -1
So, (x-4)2 + (y+1)2 = (-1)2
(x2 -8x+16) + (y2 +1+2y) = 1
x2 + y2 - 8x + 2y + 17 = 1
x2 + y2 - 8x + 2y + 16 = 0
This is the required equation of circle.

 
 The locus of the centres of the circles such that the point (2, 3) is the mid point of the chord
5x + 2y = 16 is
  • a)
     2x – 5y + 11 = 0
  • b)
    2x + 5y – 11 = 0
  • c)
    2x + 5y + 11 = 0
  • d)
    None
Correct answer is option 'A'. Can you explain this answer?

Nandini Patel answered
Let locus lie on
Y = mx +C
Slope of a line = -5/2
Hence slope of its perpendicular line = 2/5
Hence
y = 2x/5  +C
5y = 2x + 5C
(2, 3) satisfies this
C = 11/5
Hence
Line will be
5y = 2x + 5*11/5
2x - 5y + 11 = 0

The greatest distance of the point P(10, 7) from the circle x2 + y2 – 4x – 2y – 20 = 0 is
  • a)
    5
  • b)
    15
  • c)
    10
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Sankar Mehta answered
The equation of the circle is x^2 + y^2 = r^2, where r is the radius of the circle. We need to find the value of r first.

The circle passes through the origin (0,0), since x^2 + y^2 = 0^2 when x = 0 and y = 0. Therefore, the distance from the center of the circle to the origin is equal to the radius.

The center of the circle is at (0,0), so the distance from the center to the point P(10,7) is:

d = sqrt((10-0)^2 + (7-0)^2) = sqrt(149)

Therefore, the radius of the circle is r = sqrt(149).

The greatest distance from the point P(10,7) to the circle is the distance from the point to the edge of the circle along a line that passes through the center of the circle. This is equal to the difference between the distance from the center to the point and the radius of the circle.

The distance from the center of the circle to the point P(10,7) is d = sqrt((10-0)^2 + (7-0)^2) = sqrt(149).

Therefore, the greatest distance of the point P(10,7) from the circle is:

d - r = sqrt(149) - sqrt(149) = 0.

Therefore, the greatest distance is 0, which means that the point P(10,7) is on the circle.

 Pair of tangents are drawn from every point on the line 3x + 4y = 12 on the circle x2+ y2 = 4. Their variable chord of contact always passes through a fixed point whose co-ordinates are
  • a)
  • b)
  • c)
    (1, 1)
  • d)
     
Correct answer is option 'D'. Can you explain this answer?

Geetika Shah answered
Let P(x1,y1) be a point on the line 3x + 4y = 12
Equation of variable chord of contact of P(x1,y1) w.r.t circle x2 + y2 = 4 
xx1 + yy1 − 4 = 0   ...(1)
Also 3x1 + 4y1 − 12 = 0
⇒ x1 + 4/3y1 − 4 = 0   ...(2)
Comparing (1) and (2), we get
x = 1; y = 4/3
∴ Variable chord of contact always passes through (1, 4/3)

If a be the radius of a circle which touches x-axis at the origin, then its equation is
  • a)
    x2 + y2 + ax = 0
  • b)
    x2 + y2 ± 2ya = 0
  • c)
    x2 + y2 ± 2xa = 0
  • d)
    x2 + y2 + ya = 0
Correct answer is option 'B'. Can you explain this answer?

Sushant Ghosh answered
The equation of the circle  with centre at (h,k) and radius equal to a is (x−h)2 +(y−k)2 = a2
 When the circle passes through the origin  and centre lies on x− axis 
⇒h = a and k = 0
Then the equation (x−h)2+(y−k)2=a2 becomes (x−a)2+y2=a2
If a circle passes through the origin and centre lies on x−axis then the abscissa will be equal to the radius of the circle and the y−co-ordinate of the centre will be zero Hence, the equation of the circle will be of the form 
(x±a)2+y2=a2⇒x2+a2 ±2ax+y2=a2
=x2 +y2 ±2ax=0 is the required equation of the circle.

If the line 2x – y + λ = 0 is a diameter of the circle x2+y2+6x−6y+5 = 0 then λ =
  • a)
    6
  • b)
    9
  • c)
    3
  • d)
    12
Correct answer is option 'B'. Can you explain this answer?

Rajesh Gupta answered
x2 + y2 + 6x − 6y + 5 = 0
Center O = (-3, 3)
radius r = √{(-3)2 + (3)2 - 5} 
= √{9 + 9 - 5} 
= √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x – y + λ = 0
=> -3*2 - 3 + λ = 0
=> -6 - 3 + λ = 0
=> -9 + λ = 0
=> λ = 9

The equation of a circle which passes through the three points (3, 0) (1, –6), (4, –1) is
  • a)
    2x2 + 2y2 + 5x – 11y + 3 = 0
  • b)
    x2 + y2 – 5x + 11y – 3 = 0
  • c)
    x2 + y2 + 5x – 11y + 3 = 0
  • d)
    2x2 + 2y2 – 5x + 11y – 3 = 0
Correct answer is option 'D'. Can you explain this answer?

Pooja Shah answered
Gen equation of a circle is (x−h)2+(y−k)2=r2 (3,0),(1,−6),(4,−1) circle passes through these points
∴ These three points should satisfy the equation
of circle.
(3−h)2+k2=r2 ___ (I)
(1−h)2 +(−6−k)2 = r2(1−h)2+(6+k)2=r2  ___ (II)
From (I) & (II)
(3−h)2+k2 =(1−h)2+(6+k)2
9+h2−6h+k2 =1+h2 2h+36+k2+12k
9−6h=1−2h+36+12k
9−37=4h+12k
−28=4h+12k
h+3k=−7
h=−7−3k ___ (III)
(4−h)2+(−1−k)2=r2 ___ (IV)
From (I) & (IV)
(3−h)2+k2 = (4−h)2 +(1+k)2
9+h2 −6h+k2
 =16+h2 −8h+1+k2+2k
9−6h=17−8h+2k
9−17=−2h+2k
−8=−2h+2k
−h+k=−4
k=−4+h ___ (V)
Put (V) in (III)
h=−7−3(−4+h)
h=−7+12−3h
h=5/4,k=−11/4,r=170/16
Eq of circle: (x−5/4)2 +(y+11/4)2 =170/16
x2+25/16−10x/4+y2 +121/16+22y/4=170/16
Simplifying, we get 2x2+xy2−5x+11y−3=0

The number of points on X-axis which are at a distance c units (c < 3) from (2, 3) is
  • a)
    1
  • b)
    0
  • c)
    3
  • d)
    2
Correct answer is option 'B'. Can you explain this answer?

Mohit Rajpoot answered
Distance of 'c' units from (2,3)
Let the no: of points be (x,0)
By distance formula
{(2−x)2+(3−0)2}=c
 4−4x+x2+9=c
⇒x2−4x+13 = c:c=2,2
There are the points of c,such that when they are applied back to the equations,the number of points will become zero.

The equation of parabola whose focus is (– 3, 0) and directrix x + 5 = 0 is:
  • a)
    y2 = 4(x + 1)
  • b)
    y2 = 4x
  • c)
    y2 = 4(x + 16)
  • d)
    y2 = 4(x + 4)
Correct answer is option 'D'. Can you explain this answer?

Gaurav Kumar answered
Given, focus : (-3,0)
directrix : x + 5 = 0
Let (x ,y) is the point on the parabola .
∴ distance of point from focus = distance of point from directrix
⇒ √{(x + 3)² + y²} = |x + 5|/√(1² + 0²)
⇒ √{(x + 3)² + y² } = |x + 5|
squaring both sides,
(x + 3)² + y² = (x + 5)²
⇒y² = (x + 5)² - (x + 3)²
⇒y² = (x + 5 - x - 3)(x + 5 + x + 3)
⇒y² = 2(2x + 8) = 4(x + 4)
Hence, equation of parabola is y² = 4(x + 4)

The equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is:
  • a)
    x2 + y2 - 2ax - 2by = 0
  • b)
    x2 + y2 - ax - by = a2 + b2
  • c)
    x2 + y2 - ax - by = 0
  • d)
    x2 + y2 + 4x + 6y + 13 = 0
Correct answer is option 'C'. Can you explain this answer?

Varun Kapoor answered
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the centre of the circle passes through (0, 0),
(0 – h)2 + (0 – k)2 = r2
⇒ h2 + k2 = r2
The equation of the circle now becomes (x – h)2 + (y – k)2 = h2 + k2.
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,
(a – h)2 + (0 – k)2 = h2 + k2 …………… (1)
(0 – h)2 + (b – k)2 = h2 + k2 ………… (2)
From equation (1), we obtain a2 – 2ah + h2 + k2 = h2 + k2
⇒ a2 – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.
From equation (2), we obtain h2 + b2 – 2bk + k2 = h2 + k2
⇒ b2 – 2bk = 0
⇒ b(b – 2k) = 0
⇒ b = 0 or(b – 2k) = 0
However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k =b/2. Thus, the equation of the required circle is

The equation of the circle passing through (3, 6) and whose centre is (2, –1) is
  • a)
    x2 + y2 – 4x + 2y = 45
  • b)
    x2 + y2 – 4x – 2y + 45 = 0
  • c)
    x2 + y2 + 4x – 2y = 45 
  • d)
    x2 + y2 – 4x + 2y + 45 = 0
Correct answer is option 'A'. Can you explain this answer?

Meghana Pillai answered
To find the equation of the circle passing through (3, 6) and with center (2, c), we can use the distance formula.

The distance between the center (2, c) and the point (3, 6) should be equal to the radius of the circle. Let's call the radius r.

Using the distance formula:
√((3 - 2)^2 + (6 - c)^2) = r

Simplifying this equation:
√(1 + (6 - c)^2) = r

Squaring both sides of the equation:
1 + (6 - c)^2 = r^2

Now we need one more piece of information to find the value of c and r.

If the tangent at (3, –4) to the circle x2 + y2 - 4x + 2y - 5 = 0 cuts the circle x2 + y2 + 16x + 2y + 10 = 0 in A and B then the mid point of AB is
  • a)
    (-6, -7)
  • b)
    (2, -1)
  • c)
    (2, 1)
  • d)
    (5, 4)
Correct answer is option 'A'. Can you explain this answer?

Maitri Gupta answered
Solution:

Given, the equation of circle is x² + y² - 4x - 2y - 5 = 0.

Let (3, 4) be a point on the circle.

Hence, the equation of tangent at (3, 4) is

3x + 4y = 25.

We need to find the intersection points of the tangent and the circle x² + y² + 16x + 2y + 10 = 0.

Substituting y = (25 - 3x)/4 in the equation of the circle, we get

x² + ((25 - 3x)/4)² - 16x - 2((25 - 3x)/4) - 10 = 0.

Simplifying the above equation, we get

13x² - 150x + 351 = 0.

Solving this quadratic equation, we get

x = 3, 27/13.

Substituting these values of x in the equation of the tangent, we get

y = 4, -7/13.

Hence, the intersection points of the tangent and the circle are (3, 4) and (27/13, -7/13).

Let A = (3, 4) and B = (27/13, -7/13).

The midpoint of AB is (-6, -7).

Hence, the correct answer is option A.

Chapter doubts & questions for Circle - Mathematics (Maths) for JEE Main & Advanced 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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