Class 10 - Circles Archives – Mathemafia http://sh016.global.temp.domains/~mathemaf/category/class-10-circles/ Mathematics is like music - Enjoy it!!! Wed, 11 Jun 2025 09:58:35 +0000 en-US hourly 1 https://wordpress.org/?v=7.0.2 https://mathemafia.com/wp-content/uploads/2020/08/cropped-M__Logo__SM-1-32x32.png Class 10 - Circles Archives – Mathemafia http://sh016.global.temp.domains/~mathemaf/category/class-10-circles/ 32 32 Exercise 10.2 (NCERT) – Circles, Class 10 – All Solutions https://mathemafia.com/exercise-10-2-ncert-circles-class-10-all-solutions/ Wed, 11 Jun 2025 08:57:39 +0000 https://mathemafia.com/?p=19770 Complete solution of Exercise 10.2, Chapter 10, Circles of NCERT book of Mathematics for Class 10. All questions are solved with clear figures, easily readable mathematical notations along with embedded YouTube video tutorials.

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Exercise 10.2 (NCERT) - Circles, Class 10

Chapter 10 Exercise 10.2 Class 10

Below are the Quick links for all questions of Exercise 10.1, Circles, Class 10.

Click a link to view the solution of corresponding question.

Below are the solutions of every question of exercise 10.2 of chapter 10, Circles, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

Question 1

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(A) 7 cm   (B) 12 cm   (C) 15 cm   (D) 24.5 cm

Given

Length of tangent, PQ = 24 cm

OQ = 25 cm 

Let the radius = r = OP

Since PQ is a tangent to the circle at P

Hence ∠OPQ = 90

In right angled △OPQ

Using Pythagoras Theorem

Question 2

In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to

(A) 60°   (B) 70°   (C) 80°   (D) 90°

Given

∠POQ = 110°

TP and TQ are tangents from point T to the circle.

So, ∠OPT = ∠OQT = 90° (radius ⟂ tangent)

Now, in quadrilateral POQT:

  • ∠OPT = 90°
  • ∠OQT = 90°
  • ∠POQ = 110°
  • ∠PTQ = ?

Using the angle sum property of a quadrilateral:

∠PTQ = 360° – (90° + 90° + 110°) = 70°

Answer: B

Question 3

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to

(A) 50° (B) 60° (C) 70° (D) 80°

Given

∠APB = 80° 

PA and PB are tangents from point P to the circle.

OA = radius

Hence ∠OAP = 90°

Join OP

∠APB = 80°

Line OP bisects ∠APB

Therefore, ∠OPA = ∠OPB = 40°

In right angled triangle △OPA

sum of angles = 180°

So, ∠POA = 180° − 90° − 40° = 50°

Answer: A

Question 4

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Given

A circle with diameter AB. Tangents are drawn at points A and B.

To prove: The tangents at A and B are parallel.

Construction: Let O be the centre of the circle. Join OA and OB.

Since AB is the diameter => AOB is a straight line
Therefore, both tangents are perpendicular to the same line AOB.

Question 5

Prove that the perpendicular at the point of contact to the tangent to a circle passes

through the centre.

Assume that ⊥ on point P, does not pass through the centre of the circle, O.

Consider it passes through point O’.

=> ∠APO’ = 90° 

Also ∠BPO = 90° (∵ tangent is ⊥ to the radius through point of contact)

Also ∠APB = 180° (Straight angle)

This means

This is possible only when OP and O’P coincide. 

Hence our assumption is wrong.

This implies perpendicular at the point of contact to tangent passes through the centre of the circle. 

Question 6

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Given
OA = 5 cm
AP = 4 cm

To find

Radius (r) = OP

AP is tangent and OP is radius, hence

∠OPA = 90°

In right angled △OPA

Using Pythagoras Theorem

Question 7

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Let the chord AB touches the inner circle at point P. This means:

OP = 3 cm     (Radius of inner circle)

OA = 5 cm     (Radius of outer circle)

APB is tangent and OP is radius of inner circle, hence

∠OPA = 90°

In right angled △OPA

Using Pythagoras Theorem

Question 8

A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Let the circle touches sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

Let the lengths of tangents from each vertex to the points of contact be:

Using the above information

Using the above information

Question 9

In Fig. 10.13, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Watch this video to learn the easiest method to solve this question.

XY and X’Y’ are two || tangents. 

=> PAQ is a diameter of the circle.

∠1 = ∠2 = 90° 

Because AP ⊥ OP and BQ ⊥ OQ

In quadrilateral APBQ

Now, AP and AC are two tangents from the same point, A.

This implies, OA bisects ∠PAC

Similarly, OB bisects ∠QBC

In triangle AOB

Question 10

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

To Prove: ∠1 + ∠3 = 180°

Let the circle have centre O &

A be a point outside the circle.

Draw tangents AP and AQ.

Join OP and OQ.

PA is tangent and OP is radius, hence

∠OPA = ∠4 = 90°

Similarly AQ is tangent and OQ is radius, hence

∠OQA = ∠2 = 90°

Now, OPAQ is a quadrilateral.

Applying Angle sum property of quadrilateral 

∠1 + ∠2 + ∠3 + ∠4 = 360°

∠1 + 90° + ∠3 + 90° = 360°

∠1 + ∠3 + 180° = 360°

∠1 + ∠3 = 180°

Hence proved.

Question 11

Prove that the parallelogram circumscribing a circle is a rhombus.

Let ABCD be a parallelogram circumscribing a circle.

Since tangents from a point to a circle are equal:

Using above information

This means

Now AB = CD and BC = AD because ABCD is a parallelogram. Therefore

Hence ABCD is a rhombus.

Question 12

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Method 1

Given

BD = 8cm, DC = 6cm

=> BC = 14cm

Area of triangle using Heron’s Formula & inradius 

Method 2

Given

BD = 8cm, DC = 6cm

=> BC = 14cm

Area of triangle ABC = Finding areas of 3 small triangles

Area of triangle ABC by Heron’s Formula

Question 13

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

To prove

Given

ABCD is a quadrilateral circumscribing a circle.

Let centre of circle is O.

Let the circles touches the quadrilateral at points P, Q, R and S as shown in the figure.

This means APB, BQC, CRD and DSA are tangents to the circle.

Join OA, OB, OC, OD, OP, OQ, OR and OS.

Use numbers (1 to 8) to name the angles as shown in the figure.

Now we know that

Also

Similarly 

Hence proved.

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The post Exercise 10.2 (NCERT) – Circles, Class 10 – All Solutions appeared first on Mathemafia.

]]> Exercise 10.1 (NCERT) – Circles, Class 10 – All Solutions https://mathemafia.com/exercise-10-1-ncert-circles-class-10-all-solutions/ Mon, 09 Jun 2025 07:46:09 +0000 https://mathemafia.com/?p=19698 Complete solution of Exercise 10.1, Chapter 10, Circles of NCERT book of Mathematics for Class 10. The exercise is very easy and short, primarily about some properties of circle.

The post Exercise 10.1 (NCERT) – Circles, Class 10 – All Solutions appeared first on Mathemafia.

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Exercise 10.1 (NCERT) - Circles, Class 10

Chapter 10 Exercise 10.1 Class 10

Below are the Quick links for all questions of Exercise 10.1, Circles, Class 10.

Click a link to view the solution of corresponding question.

Below are the solutions of every question of exercise 10.1 of chapter 10, Circles, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

Question 1

How many tangents can a circle have?

Infinitely many. 

 

Reason: There are infinitely many points on the circumference of a circle. Hence infinitely many tangents. 

Still confused? Watch the video.

Question 2

Fill in the blanks :

(i) A tangent to a circle intersects it in 1 point (s).

(iii) A line intersecting a circle in two points is called a secant.

(ii) A circle can have 2 parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called point of contact.

Question 3

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :

(A) 12 cm (B) 13 cm (C) 8.5 cm (D) √119 cm.

OP = 5 cm (radius)

OQ = 12 cm

Since PQ is a tangent to the circle at P

∠OPQ = 90

In right triangle OPQ, apply the Pythagoras Theorem

Question 4

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

You need to watch the video 😉

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The post Exercise 10.1 (NCERT) – Circles, Class 10 – All Solutions appeared first on Mathemafia.

]]> Class 10 | Chapter 10 | All Solutions | Circles https://mathemafia.com/class-10-chapter-10-all-solutions-circles/ Fri, 17 Dec 2021 07:41:58 +0000 http://sh016.global.temp.domains/~mathemaf/?p=7355 This playlist covers all the questions, concepts and examples of chapter 10 (Circles) of class 10, NCERT book. Click a thumbnail to watch the tutorial.  Watch complete chapter on YouTube

The post Class 10 | Chapter 10 | All Solutions | Circles appeared first on Mathemafia.

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This playlist covers all the questions, concepts and examples of chapter 10 (Circles) of class 10, NCERT book. Click a thumbnail to watch the tutorial. 

Watch complete chapter on YouTube

The post Class 10 | Chapter 10 | All Solutions | Circles appeared first on Mathemafia.

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