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Circles questions are a major part of geometry asked in competitive exams. These questions carry a weightage of 1-2 questions (2-4 marks) in SSC exams. To get a good rank in competitive exams, you should have a conceptual clarity of questions of circles as they also are part of mixed geometry questions asked.

Here are some tips for solving Circles questions: Grasp circle fundamentals. Learn geometry rules for tangents, secants, and chords. Understand inscribed angles and central angles, and practice accurate diagram drawing for visualization. Utilize Pythagoras&#39; theorem in right triangles associated with circles.

So, we have attached 10 questions of Circles for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Circles</h2>

You can also download the Circles questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 In the given figure, O is the centre of the circle. If ∠AOB = 88° then what is the value of ∠ACB?

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-406488,.png" />

1. 130°

2. 132°

3. 134°

4. 136°

(Difficulty: 3, Estimated Time: 20 Seconds) A good question to start with....

Q:2 In the given figure, QS and RS are two tangents to the circle with center O. If ∠QSR = 80° then what is the value of ∠QPR?

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-406477,.png" />

1. 40°

2. 45°

3. 50°

4. 55°

(Difficulty: 3, Estimated Time: 20 Seconds) You won&#39;t get easy questions in exam...

Q:3 In the given figure, QT is tangent to the circle with radius 5 cm. If T is at a distance of 11 cm from the centre O and TR is 6 cm, then what is the length of chord PR?

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-406474,.png" />

1. 5 cm 

2. 10 cm

3. 15 cm 

4. 20 cm

(Difficulty: 3, Estimated Time: 20 Seconds) Do you remember the relation.....

Q:4 In the given figure, if a = 70° then what is the value of b?

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-406358,.png" />

1. 20°

2. 30°

3. 40°

4. 50°

(Difficulty: 2, Estimated Time: 15 Seconds) This was quite easy.....

Q:5 in the given figure O is the center of the circle and ∠XPY = 30°. Find the average of the ∠XOY and ∠ XZY.

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-407619,.png" style="height:129px; width:260px" />

1. 150°

2. 255°

3. 127.5°

4. 105°

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 MN and QR are two chords in a circle with center O and MR is the circle&#39;s diameter. When MN and QR are further extended, it meets at a point P outside the Circle. if the ∠RMN = 38°, ∠MPR = 32° then find the angle ∠QNR.

1. 35°

2. 25°

3. 30°

4. 20°

(Difficulty: 4, Estimated Time: 25 Seconds) You should have a good practice to solve such.....

Q:7 PQ and MN are the two chords that intersect on point O in the circle. If the length of MN  is 11cm, PO is 4 cm, and QO is 7.5 cm then, find the difference between the ON and OM.

1. 1

2. 2

3. 3

4. 4

(Difficulty: 3, Estimated Time: 20 Seconds) Be prepared for such questions in exam!

Q:8 If the radii of the two circles, not touching each other, are 8 cm and 7 cm respectively and the distance between the centres of the circles is 25 cm then find the length of the common transversal tangent. 

1. 15 cm

2. 20 cm

3. 25 cm

4. 30 cm

(Difficulty: 2, Estimated Time: 15 Seconds) This was not a hard question!

Q:9 If two circles with radius 25 cm and 16 cm respectively touch each other externally, then find the length of the common tangent to both the circles.

1. 20 cm

2. 30 cm

3. 40 cm

4. 50 cm

(Difficulty: 3, Estimated Time: 20 Seconds) A similar one....

Q:10 In the given figure, what will be the value of x (in cm)?

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-407298,.png" style="height:131px; width:226px" />

1. 1

2. 2

3. 3

4. 4

(Difficulty: 3, Estimated Time: 20 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (4)</td>
			<td>2. (3)</td>
			<td>3. (2)</td>
			<td>4. (1)</td>
			<td>5. (3)</td>
		</tr>
		<tr>
			<td>6. (4)</td>
			<td>7. (1)</td>
			<td>8. (2)</td>
			<td>9. (3)</td>
			<td>10. (4)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 4 i.e. 136°.

Given:

∠AOB = 88°

Concept Used:

Central angle is double the angle of the major arc.

The sum of the opposite angles of a cyclic quadrilateral is 180°.

Explanation:

In the given figure,

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent Diagram_05.06.2023_D2.png" style="height:150px; width:147px" />

The lines from A and B touches the circumference of the circle at point D, such that ∠ADB = ½ × ∠AOB 

∠AOB = 88° 

Then, ∠ADB = ½ × 88° 

⇒ ∠ADB = 44°

Now, In quadrilateral ABCD, ∠C + ∠D = 180°

⇒ ∠C + 44° = 180°

⇒ ∠C = 180° - 44°

⇒ ∠C = 136°

Hence the value of ∠ACB is 136°.

 

Q:2 The correct answer is option 3 i.e. 50°

Given:

∠QSR = 80°

Concept Used:

The angle between the radius of the circle and the tangent to the circle is 90°.

Sum of all angle of a quadrilateral is 360°.

The angle on the circumference of the circle is half the angle at the center of the circle, when subtended from the same arc.

Explanation:

In the figue given,

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/(Ans) HRQ-406477,.png" />

∠Q = ∠R = 90°

Now, in quadrilateral QORS, ∠Q + ∠O + ∠R + ∠S = 360°

⇒ 90° + ∠O + 90° + 80° = 360°

⇒ ∠O + 260° = 360°

⇒ ∠O = 100°

Now, ∠QPR = ½ × ∠O

⇒ ∠QPR = ½ × 100°

⇒ ∠QPR = 50°

Hence the value of ∠QPR is 50°.

 

Q:3 The correct answer is option 2 i.e. 10 cm.

Given:

OT = 11 cm

Radius, OC = 5 cm

TR = 6 cm

Concept Used:

If any two chords, say AB and CD, of a circle, intersect each other at a point E outside a circle, then AE × EB = EC × ED

Explanation:

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/(Ans) HRQ-406474,.png" />

In the above figure, TA = OT + OA = 11 + 5 = 16 cm

TC = TA - AC = 16 - 10 = 6 cm

Now, TR × TP = TC × TA

⇒ 6 × TP = 6 × 16

⇒ TP = 16 cm

So, PR = TP - TR

⇒ PR = 16 - 6 

⇒ PR = 10 cm

Hence the length of PR is 10 cm

 

Q:4 The correct answer is option 1 i.e. 20°.

Given:

a = 70°

Concept Used:

The tangent to the circle makes a right angle with tha radius of the circle.

Sum of all angles of a triangle is 180°.

Explanation:

∠P = ∠Q = 90°  {PR and QR are tangents to the circle}

Now, in triangle POR,

∠P + ∠O + ∠R = 180°

⇒ 90° + 70° + ∠R = 180°

⇒ ∠R = 180° - 160°

⇒ ∠R = 20°

Hence the value of b is 20°.

 

Q:5 The correct answer is option 3 i.e. 127.5°.

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-407619,.png" style="height:109px; width:220px" />

⇒ ∠XPY = 30°

⇒ ∠OXP = ∠OYZ = 90° (Tangent on a circle)

⇒ ∠YOX + ∠OXP + ∠OYZ + ∠XPY = 360°

⇒ ∠YOX = 360° - (90° + 90° + 30) = 150°

⇒ ∠XZY = 90° + ∠XPY/2 = 90 + 15 = 105°

Required average = (150° + 105°)/2 = 127.5°

 

Q:6 The correct answer is option 4 i.e. 20°.

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-407613,.png" style="height:119px; width:220px" />

∠RMN = 38°, ∠MPR = 32° Given

∠MNR = 90° (The angle subtended in a semicircle is a right angle)

∠MRQ = 32° + 38° = 70° (The exterior angle of the given triangle equals to the sum of the two opposite interior angles present in that triangle)

∠MRQ = ∠QNM = 70°

∠QNR = 90° - 70° = 20°

 

Q:7 The correct answer is option 2 i.e.1.

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent Diagram_05.06.2023_D5.png" />

We know that if the two chords in a circle intersect each other on point O then,

⇒ OP × OQ = OM × ON

⇒ 4 × 7.5 = x(11 - x)

⇒ 30 = 11x - x2

⇒ x2 - 11x + 30 = 0

⇒ x = 6, 5

If x = 6,

Then MO = 6, NO = (11 - 6) = 5

So, the difference = (6 - 5) = 1

 

Q:8 The correct answer is option 2 i.e.  20 cm.

Formula Used:

AB2 = (OO&#39;)2 - (OA + O&#39;B)2

Explanation:

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent Diagram_05.06.2023_D6.png" />

From the formula, we get

AB2 = (25)2 - (8 + 7)2

⇒ AB2 = 625 - 225

⇒ AB2 = 400

⇒ AB = \(\sqrt{400}\)

⇒ AB = 20

Hence, the value of a common transverse tangent is 20 cm

 

Q:9 The correct answer is option 3 i.e. 40 cm.

Given:

R1 = 25 cm

R2 = 16 cm

Formula Used:

(Length of common tangent)2 = (Distance between the centers of the circle)2 - (R1 - R2)2 

Explanation:

In the given question, the two circles touch each other externally, then the distance between their centres is equal to the sum of their radii.

Now, according to the formula

(Common Tangent)2 = (25 + 16)2 - (25 - 16)2

⇒ (Common Tangent)2 = (41)2 - (9)2

⇒ (Common Tangent)2 = (41 + 9)(41 - 9)

⇒ (Common Tangent)2 = 50 × 32

⇒ (Common Tangent)2 = 1600

⇒ Common tangent = \(\sqrt{1600}\)

⇒ Common Tangent = 40

Hence the value of common tangent is 40 cm.

 

Q:10 The correct answer is option 4 i.e. 4 cm.

Given :

AD = 7 cm

BD = 5 cm

BE = 6 cm

Concept Used:

If the two chords, say AB and CD, of a circle, intersect each other at point E outside the circle, then 

EA × EB = EC × ED

Explanation:

In the given question,

BD × BA = BE × BC

⇒ 5 × (7 + 5) = 6 × (6 + x)

⇒ 5 × 12 = 6 × (6 + x)

⇒ 6 + x = 10

⇒ x = 4

Hence, the value of x is 4 cm

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Circles.

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