Circles MCQ Quiz - Objective Question with Answer for Circles - Download Free PDF

Last updated on Jul 4, 2025

Circles MCQs put a spotlight on fundamental geometry concepts involving circles. Topics covered include their properties, equations, tangents, secants, and various theorems. Circles constitute a significant part of geometry and mastering them provides an edge in competitive mathematics exams. These MCQs provide a thorough grounding in the subject and help students develop both speed and accuracy in solving problems. Solidify your understanding of circles, boost your preparation by solving Circles MCQs with answers right away.

Latest Circles MCQ Objective Questions

Circles Question 1:

A square is inscribed in a circle x 2 + y 2 + 2x + 2y + 1 = 0 and its sides are parallel to coordinate axes. Which one of the following is a vertex of the square?

  1. (-2,2)
  2. (-2,-2)
  3. (1+12,112)
  4. None of the above

Answer (Detailed Solution Below)

Option 3 : (1+12,112)

Circles Question 1 Detailed Solution

Calculation:

qImage6842888393ed0e6eef0d3d21

Given,

The circle’s equation is

x2+y2+2x+2y+1=0

Rewrite by completing squares:

(x+1)2+(y+1)2=1

∴ center = (-1, -1), radius r = 1

For a square inscribed in this circle with sides parallel to the axes, its diagonal equals the circle’s diameter = 2. If the square has side length a, then

a2=2a=2.

Each vertex lies a half‐side =a2=12 from the center along both axes. Since the center is ( -1, -1), the four vertices are

(1±12,1±12).

One of these vertices is

(1+12,112).

Hence, the correct answer is Option 3

Circles Question 2:

What is the equation of the circle whose diameter is 10 cm and the equations of two of its diameters are and ?

  1. x2+y2=1
  2. x2+y2=25
  3. x2+y2=100
  4. x2+y22x2y23=0

Answer (Detailed Solution Below)

Option 2 : x2+y2=25

Circles Question 2 Detailed Solution

Calculation:

Given two diameters of the circle are along the lines x + y = 0 and x - y = 0. The center of the circle is at their intersection.

We have, 

xy=0x=y.

Substitute into x + y = 0:

x+x=02x=0x=0,y=0.

Thus, the center is (0, 0).

Also radius,

r=102=5.

the equation of the circle with center (0,0) and radius 5 is

x2+y2=52=25.

Hence, the correct answer is Option 2.

Circles Question 3:

Points P(–3, 2),Q(9, 10) and R(α, 4)lie on a circle C with PR as its diameter. The tangents to C at the points Q and R intersect at the point S. If S lies on the line 2x – ky = 1, then k is equal to _____. 

Answer (Detailed Solution Below) 3

Circles Question 3 Detailed Solution

Calculation: 

qImage683545bdbc2acb4bbe3db33b

mPQ · mQR = –1

1029+3×1049α=1α=13

m0PmQS=1mQS=47

Equation of QS 

y10=47(x9)

⇒ 4x + 7y = 106 ….(1)

m0R · mRS= –1 mRS = – 8

Equation of RS

y – 4 = –8(x–13)

⇒ 8x + y = 108 ….(2)

Solving eq. (1) & (2)

⇒ x1=252y1=8

S(x1, y1) lies on 2x – ky = 1

⇒ 25 – 8k = 1

⇒ 8k = 24

 k = 3

Hence, the correct answer is 3. 

Circles Question 4:

The distance of the point (6, -2√2) − from the common tangent y = mx + c, m > 0, of the curves x = 2y2 and x = 1 + y2 is 

  1. 13
  2. 5
  3. 143
  4. 5√3

Answer (Detailed Solution Below)

Option 2 : 5

Circles Question 4 Detailed Solution

Calculation: 

For y2=x2,T:y=mx+18m

For tangent to y2 + 1 = x

⇒ (mx+18m)2+1=x

D=0m=122

∴ T : x - 2√2y + 1 = 0

⇒ d=|6+8+19|=5

Hence, the correct answer is Option 2.

Circles Question 5:

If the four distinct points (4, 6), (–1, 5), (0, 0) and (k, 3k) lie on a circle of radius r, then 10k + r2 is equal to 

  1. 32 
  2. 33 
  3. 34 
  4. 35

Answer (Detailed Solution Below)

Option 4 : 35

Circles Question 5 Detailed Solution

Calculation: 

qImage6825a7b87646fb13ad1ad8bc

m1m2 = – 1 so right angle equation circle is

⇒ (x – 4) (x – 0) + (y – 6) (y – 0) = 0

⇒ x2 + y2 – 4x – 6y = 0

⇒ (k,3k) lies on it so

⇒ k2 + 9k2 – 4k – 18k = 0

⇒ 10k2 – 22k = 0 

⇒ k = 0, 115

k = 0 is not possible so k = 115

also r = 4+9=13

so 10k + r2 10. 115+(13)2=35

Hence, the Correct answer is Option 4.

Top Circles MCQ Objective Questions

AB is a chord in the minor segment of a circle with center O. C is a point between A and B on the minor arc AB. The tangents to the circle at A and B meet at the point D. If ∠ACB = 116°, then the measure of ∠ADB is

  1. 64°
  2. 52°
  3. 56°
  4. 48°

Answer (Detailed Solution Below)

Option 2 : 52°

Circles Question 6 Detailed Solution

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Given:

F4 Savita SSC 16-11-22 D1 V2

∠ACB = 116°

Concept Used:

When a quadrilateral is inscribed in a circle, the opposite angles of it are supplementary angles.

The angle subtended at the center is always double the angle subtended at the remaining arc.

The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.

Calculation:

Point P is taken on the major arc of the circle.

Then, A & P, B & P, C & B, and A & C are joined.

From cyclic quadrilateral APBC

∠ACB = 116°

Now, ∠APB = (180 – 116)° = 64°

Now, ∠AOB = (64 × 2)°

⇒ 128°

Since OA = OB = radius of the circle

So, from quadrilateral AOBD

∠ADB = 360° - (∠OBD + ∠OAD + ∠AOB)

⇒ ∠ADB = 360° - (90° + 90° + 128°)

⇒ ∠ADB = 360° - 308°

⇒ ⇒ ∠ADB = 52° 

∴ The required measure of ∠ADB is 52°.

The radius of the circle 2x2 + 2y2 + 8x + 8y + 4 = 0 is

  1. 2 unit 
  2. 4 unit
  3. √6 unit
  4. 5 unit

Answer (Detailed Solution Below)

Option 3 : √6 unit

Circles Question 7 Detailed Solution

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Concept:

The general second degree equation of a circle in x and y is given by:  x2 + y2 + 2gx + 2fy + c = 0 with centre (-g, -f) and radius

 r=g2+f2c

 

Calculation:

Given:   2x2 + 2y2 + 8x + 8y + 4 = 0 

⇒ 2 × (x2 + y2 + 4x + 4y + 2) = 0

⇒ x2 + y2 + 4x + 4y + 2 = 0 are equation of circle with centres C and radius r 

By comparing the equation of the circle with the equation  x 2 + y2 + 2gx + 2fy + c = 0 we get 

g = 2, f =  2 and c = 2

As we know, radius = r=g2+f2c

r=4+42

r = √6 unit

In a circle with centre O, a 6 cm long chord is at a distance 4 cm from the centre. Find the length of the diameter.

  1. 5 cm
  2. 10 cm
  3. 14 cm
  4. 7 cm

Answer (Detailed Solution Below)

Option 2 : 10 cm

Circles Question 8 Detailed Solution

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Concept: 

Any line from the center that bisects a chord is perpendicular to the chord.

Pythagoras theorem: 

h2 = b2 + p2

Calculation:

F1 Harshit 15-09-21 Savita D4

r2 = 4+ 32

⇒ r = 5

∴ The diameter of the circle = 2r = 2 × 5

⇒ 10 cm 

Shortcut TrickBy triplets:

3, 4 and 5 

The length of the diameter = 2× 5 = 10

The circle x2+y2+4x7y+12=0 cuts an intercept on y-axis of length

  1. 3
  2. 4
  3. 7
  4. 1

Answer (Detailed Solution Below)

Option 4 : 1

Circles Question 9 Detailed Solution

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Concept:

On X- axis y-intercept will be zero, similarly, on Y-axis x-intercept will be zero.

 

Calculation:

On Y-axis x-intercept = 0

So, 

x2+y2+4x7y+12=0y27y+12=0y24y3y+12=0y(y4)3(y4)=0(y3)(y4)=0

 y = 3 and 4

Points (0, 3) and (0, 4)

Now length =

=02+(34)2=1

Hence, option (4) is correct.

AB is the diameter of a circle with centre O. C and D are two points on the circle on either side of AB, such that ∠CAB = 52° and ∠ABD = 47°. What is the difference (in degrees) between the measures of ∠CAD and ∠CBD?

  1. 25
  2. 15
  3. 10
  4. 20

Answer (Detailed Solution Below)

Option 3 : 10

Circles Question 10 Detailed Solution

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Given:

∠CAB = 52° and ∠ABD = 47°

Concept used:

The angle subtended by the diameter at any point on the circumference is equal to 90°

Calculation:

F2 Savita SSC 7-9-22 D2

According to the concept,

∠ACB = ∠ADB = 90°

So, ∠CBA = 180° - (90° + 52°)

⇒ 180° - 142°

⇒ 38°

Similarly,

∠BAD = 180° - (90° + 47°)

⇒ 180° - 137°

⇒ 43°

So, ∠CAD = 52° + 43°

⇒ 95°

∠CBD = 38° + 47°

⇒ 85°

Now,

∠CAD - ∠CBD = 95° - 85°

⇒ 10°

∴ The difference (in degrees) between the measures of ∠CAD and ∠CBD is 10.

The sum of the lengths of the radius and the diameter of a circle is 84 cm. What is the difference between the lengths of the circumference and the radius of this circle? [Use π=227]

  1. 156 cm
  2. 172 cm
  3. 148 cm
  4. 128 cm

Answer (Detailed Solution Below)

Option 3 : 148 cm

Circles Question 11 Detailed Solution

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Given:

Radius + Diameter = 84 cm

Formula used:

Diameter = 2 x Radius

Circumference = 2πr

Calculations:

Let the radius be R then, Diameter be 2R.

Now, according to question, 

R + 2R = 84

⇒ 3R = 84

⇒ R = 28 

So, The circumference = 2πr = 2 × 227 × 28 = 176 cm

Now, difference between Circumference and radius is

= 176 - 28

= 148 cm

Hence, the required difference is 148 cm.

What is the length of the tangent to the circle x2 + y2 = 9 from the point (4, 0).

  1. √7 units
  2. √6 units
  3. √11 units
  4. √17 units

Answer (Detailed Solution Below)

Option 1 : √7 units

Circles Question 12 Detailed Solution

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Concept:

The length of the tangent from an external point (x1, y1) to the circle x2 + y2 = a2 is x12+y12a2

Calculation:

Given: Equation of circle x2 + y2 = 9 and the point (4, 0).

As we know that, the length of the tangent from an external point (x1, y1) to the circle x2 + y2 = a2 is x12+y12a2

Here, x1 = 4 , y1 = 0 and a2 = 9.

So, the length of the tangent is √7 units.

Find the equation of the circle whose centre is at (2, - 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4 ?

  1. x2 + y2 - 4x + 6y + 3 = 0
  2. x2 + y2 + 4x + 6y + 3 = 0
  3. x2 + y2 - 4x + 6y - 3 = 0
  4. None of these

Answer (Detailed Solution Below)

Option 1 : x2 + y2 - 4x + 6y + 3 = 0

Circles Question 13 Detailed Solution

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CONCEPT:

Equation of circle with centre at (h, k) and radius r units is given by: (x - h)2 + (y - k)2 = r2

CALCULATION:

Here, we have to find the equation of the circle whose centre is at (2, - 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4

First let's find the point of intersection of the lines  3x + 2y = 11 and 2x + 3y = 4

So, by solving the equations  3x + 2y = 11 and 2x + 3y = 4, we get x = 5 and y = - 2

So, the required circle passes through the point (5, - 2)

Let the radius of the required circle be r

As we know that, the equation of circle with centre at (h, k) and radius r units is given by: (x - h)2 + (y - k)2 = r2

Here, we have h = 2 and k = - 3

⇒ (x - 2)2 + (y + 3)2 = r2 ------------(1)

∵ The circle  passes through the point (5, - 2)

So, x = 5 and y = - 2 will satisfy the equation (1)

⇒ (5 - 2)2 + (- 2 + 3)2 = r2

⇒ r2 = 10

So, the equation of the required circle is (x - 2)2 + (y + 3)2 = 10

⇒ x2 + y2 - 4x + 6y + 3 = 0

So, the equation of the required circle is x2 + y2 - 4x + 6y + 3 = 0

Hence, option A is the correct answer.

The equation of circle with centre (1, -2) and radius 4 cm is:

  1. x2 + y2 + 2x - 4y = 16
  2. x2 + y2 + 2x - 4y = 11
  3. x2 + y2 + 2x + 4y = 16
  4. x2 + y2 - 2x + 4y = 11

Answer (Detailed Solution Below)

Option 4 : x2 + y2 - 2x + 4y = 11

Circles Question 14 Detailed Solution

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Given

Centre point are  (1, -2)

Radius = 4cm

Formula used

(x -a)2 + (y - b)2 = r2

where, a and b are point on centre

r = radius

x and y be any point on circle

Calculation

F9 Aman Kumar 9-2-2021 Swati D7 

 

Put the value of  a, b and r in the formula

(x-1)2 + (y + 2)2  = 16

⇒ x+ 1 - 2x + y2 + 4 + 4y = 16

⇒ x2 + y2 - 2x + 4y = 11

The radius of the circle x2 + y2 + x + c = 0 passing through the origin is

  1. 1 / 4
  2. 1 / 2
  3. 1
  4. 2

Answer (Detailed Solution Below)

Option 2 : 1 / 2

Circles Question 15 Detailed Solution

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Concept:

Let x2 + y2 = r2 is the equation of circle. then (0, 0) is the origin and r is the radius of the circle. 

Calculations:

We know that, x2 + y2 = r2 is the equation of circle. then (0, 0) is the origin and r is the radius of the circle.

Given equation of circle is x2 + y2 + x + c = 0, which is passing through the origin.

i.e. c = 0.

⇒x2 + y2 + x = 0.

⇒ x2 + x + 1414 + y2 = 0

⇒x2 + x + 14+ y14

(x+12)2+y2=(12)2

which is equation of circle with radius is 12.

Hence, the radius of the circle x2 + y2 + x + c = 0 passing through the origin is 12.

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