# NCERT Solutions For Class 10 Maths Chapter 10 Circles

**NCERT Solutions for Class 10 Maths Chapter 10 Circles** are prepared after thorough research by highly experienced Maths teachers, at BYJUโS. This study material is very important for your Class 10 board exam preparation. We have provided step by step answers to all the questions provided in the **NCERT class 10 Maths textbook.** This solution is free to download and the questions are systematically arranged for your ease of preparation and in solving different types of questions. In order to score good marks, students are advised to learn these NCERT solution.Chapter 10 Circles

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### Access Answers of Maths NCERT solutions for class 10 Chapter 10 โ Circles

**Exercise: 10.1 (Page No: 209)**

**1. How many tangents can a circle have?**

**Answer:**

There can be **infinite** tangents to a circle. A circle is made up of infinite points which are at an equal distance from a point. Since there are infinite points on the circumference of a circle, infinite tangents can be drawn from them.

**2. Fill in the blanks:**

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

**(ii) A line intersecting a circle in two points is called a โฆโฆโฆโฆ.**

**(iii) A circle can have โฆโฆโฆโฆโฆ parallel tangents at the most.**

**(iv) The common point of a tangent to a circle and the circle is called โฆโฆโฆโฆ**

**Answer:**

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

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

(iii) A circle can have **two **parallel tangents at the most.

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

**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) โ119 cm**

**Answer:**

In the above figure, the line that is drawn from the centre of the given circle to the tangent PQ is perpendicular to PQ.

And so, OP โฅ PQ

Using Pythagoras theorem in triangle ฮOPQ we get,

OQ^{2} = OP^{2}+PQ^{2}

(12)^{2 }= 5^{2}+PQ^{2}

PQ^{2} = 144-25

PQ^{2} = 119

PQ = โ119 cm

So, **option D** i.e. โ119 cm is the length of PQ.

**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.**

**Answer:**

In the above figure, XY and AB are two the parallel lines. The line segment AB is the tangent at point C while the line segment XY is the secant.

**Exercise: 10.2 (Page NO: 213)**

**In Q.1 to 3, choose the correct option and give justification.**

**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**

**Answer:**

First, draw a perpendicular from the center O of the triangle to a point P on the circle which is touching the tangent. This line will be perpendicular to the tangent of the circle.

So, OP is perpendicular to PQ i.e. OP โฅ PQ

From the above figure, it is also seen that โณOPQ is a right angled triangle.

It is given that

OQ = 25 cm and PQ = 24 cm

By using Pythagoras theorem in โณOPQ,

OQ^{2} = OP^{2} +PQ^{2}

(25)^{2 }= OP^{2}+(24)^{2}

OP^{2} = 625-576

OP^{2} = 49

OP = 7 cm

So, option A i.e. 7 cm is the radius of the given circle.

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

**(A) 60ยฐ**

**(B) 70ยฐ**

**(C) 80ยฐ**

**(D) 90ยฐ**

**Answer:**

From the question, it is clear that OP is the radius of the circle to the tangent PT and OQ is the radius to the tangents TQ.

So, OP โฅ PT and TQ โฅ OQ

โดโ OPT = โ OQT = 90ยฐ

Now, in the quadrilateral POQT, we know that the sum of the interior angles is 360ยฐ

So, โ PTQ+โ POQ+โ OPT+โ OQT = 360ยฐ

Now, by putting the respective values we get,

โ PTQ +90ยฐ+110ยฐ+90ยฐ = 360ยฐ

โ PTQ = 70ยฐ

So, โ PTQ is 70ยฐ which is option B.

**3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80ยฐ, then โ POA is equal to**

**(A) 50ยฐ**

**(B) 60ยฐ**

**(C) 70ยฐ**

**(D) 80ยฐ**

**Answer:**

First, draw the diagram according to the given statement.

Now, in the above diagram, OA is the radius to tangent PA and OB is the radius to tangents PB.

So, OA is perpendicular to PA and OB is perpendicular to PB i.e. OA โฅ PA and OB โฅ PB

So, โ OBP = โ OAP = 90ยฐ

Now, in the quadrilateral AOBP,

The sum of all the interior angles will be 360ยฐ

So, โ AOB+โ OAP+โ OBP+โ APB = 360ยฐ

Putting their values, we get,

โ AOB + 260ยฐ = 360ยฐ

โ AOB = 100ยฐ

Now, consider the triangles โณOPB and โณOPA. Here,

AP = BP (Since the tangents from a point are always equal)

OA = OB (Which are the radii of the circle)

OP = OP (It is the common side)

Now, we can say that triangles OPB and OPA are similar using SSS congruency.

โดโณOPB โ โณOPA

So, โ POB = โ POA

โ AOB = โ POA+โ POB

2 (โ POA) = โ AOB

By putting the respective values, we get,

=>โ POA = 100ยฐ/2 = 50ยฐ

As angle โ POA is 50ยฐ option A is the correct option.

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

**Answer:**

First, draw a circle and connect two points A and B such that AB becomes the diameter of the circle. Now, draw two tangents PQ and RS at points A and B respectively.

Now, both radii i.e. AO and OP are perpendicular to the tangents.

So, OB is perpendicular to RS and OA perpendicular to PQ

So, โ OAP = โ OAQ = โ OBR = โ OBS = 90ยฐ

From the above figure, angles OBR and OAQ are alternate interior angles.

Also, โ OBR = โ OAQ and โ OBS = โ OAP (Since they are also alternate interior angles)

So, it can be said that line PQ and the line RS will be parallel to each other. (Hence Proved).

**5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.**

**Solution:**

First, draw a circle with center O and draw a tangent AB which touches the radius of the circle at point P.

**To Proof:** PQ passes through point O.

Now, let us consider that PQ doesnโt pass through point O. Also, draw a CD parallel to AB through O. Here, CD is a straight line and AB is the tangent. Refer the diagram now.

From the above diagram, PQ intersects CD and AB at R and P respectively.

AS, CD โฅ AB,

Here, the line segment PQ is the line of intersection.

Now angles ORP and RPA are equal as they are alternate interior angles

So, โ ORP = โ RPA

And,

โ RPA = 90ยฐ (Since, PQ is perpendicular to AB)

โ ORP = 90ยฐ

Now, โ ROP+โ OPA = 180ยฐ (Since they are co-interior angles)

โ ROP+90ยฐ = 180ยฐ

โ ROP = 90ยฐ

Now, it is seen that the โณORP has two right angles which are โ ORP and โ ROP. Since this condition is impossible, it can be said the supposition we took is wrong.

**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.**

**Answer:**

Draw the diagram as shown below.

Here, AB is the tangent that is drawn on the circle from a point A.

So, the radius OB will be perpendicular to AB i.e. OB โฅ AB

We know, OA = 5cm and AB = 4 cm

Now, In โณABO,

OA^{2} =AB^{2}+BO^{2 }(Using Pythagoras theorem)

5^{2 }= 4^{2}+BO^{2}

BO^{2} = 25-16

BO^{2} = 9

BO = 3

So, the radius of the given circle i.e. BO is 3 cm.

**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.**

**Answer:**

Draw two concentric circles with the center O. Now, draw a chord AB in the larger circle which touches the smaller circle at a point P as shown in the figure below.

From the above diagram, AB is tangent to the smaller circle to point P.

โด OP โฅ AB

Using Pythagoras theorem in triangle OPA,

OA^{2}= AP^{2}+OP^{2}

5^{2} = AP^{2}+3^{2}

AP^{2} = 25-9

AP = 4

Now, as OP โฅ AB,

Since the perpendicular from the center of the circle bisects the chord, AP will be equal to PB

So, AB = 2AP = 2ร4 = 8 cm

So, the length of the chord of the larger circle is 8 cm.

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

**Answer:**

The figure given is:

From this figure we can conclude a few points which are:

(i) DR = DS

(ii) BP = BQ

(iii) AP = AS

(iv) CR = CQ

Since they are tangents on the circle from points D, B, A, and C respectively.

Now, adding the LHS and RHS of the above equations we get,

DR+BP+AP+CR = DS+BQ+AS+CQ

By rearranging them we get,

(DR+CR) + (BP+AP) = (CQ+BQ) + (DS+AS)

By simplifying,

AD+BC= CD+AB

**9. In Fig. 10.13, XY and XโฒYโฒ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XโฒYโฒ at B. Prove that โ AOB = 90ยฐ.**

**Answer:**

From the figure given in the textbook, join OC. Now, the diagram will be as-

Now the triangles โณOPA and โณOCA are similar using SSS congruency as:

(i) OP = OC They are the radii of the same circle

(ii) AO = AO It is the common side

(iii) AP = AC These are the tangents from point A

So, โณOPA โ โณOCA

Similarly,

โณOQB โ โณOCB

So,

โ POA = โ COA โฆ (Equation i)

And, โ QOB = โ COB โฆ (Equation ii)

Since the line POQ is a straight line, it can be considered as a diameter of the circle.

So, โ POA +โ COA +โ COB +โ QOB = 180ยฐ

Now, from equations (i) and equation (ii) we get,

2โ COA+2โ COB = 180ยฐ

โ COA+โ COB = 90ยฐ

โดโ AOB = 90ยฐ

**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 center.**

**Answer:**

First, draw a circle with centre O. Choose an external point P and draw two tangents PA and PB at point A and point B respectively. Now, join A and B to make AB in a way that it subtends โ AOB at the center of the circle. The diagram is as follows:

From the above diagram, it is seen that the line segments OA and PA are perpendicular.

So, โ OAP = 90ยฐ

In a similar way, the line segments OB โฅ PB and so, โ OBP = 90ยฐ

Now, in the quadrilateral OAPB,

โดโ APB+โ OAP +โ PBO +โ BOA = 360ยฐ (since the sum of all interior angles will be 360ยฐ)

By putting the values we get,

โ APB + 180ยฐ + โ BOA = 360ยฐ

So, โ APB + โ BOA = 180ยฐ (Hence proved).

**11. Prove that the parallelogram circumscribing a circle is a rhombus.**

**Answer:**

Consider a parallelogram ABCD which is circumscribing a circle with a center O. Now, since ABCD is a parallelogram, AB = CD and BC = AD.

From the above figure, it is seen that,

(i) DR = DS

(ii) BP = BQ

(iii) CR = CQ

(iv) AP = AS

These are the tangents to the circle at D, B, C, and A respectively.

Adding all these we get,

DR+BP+CR+AP = DS+BQ+CQ+AS

By rearranging them we get,

(BP+AP)+(DR+CR) = (CQ+BQ)+(DS+AS)

Again by rearranging them we get,

AB+CD = BC+AD

Now, since AB = CD and BC = AD, the above equation becomes

2AB = 2BC

โด AB = BC

Since AB = BC = CD = DA, it can be said that ABCD is a rhombus.

**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.**

**Answer:**

The figure given is as follows:

Consider the triangle ABC,

We know that the length of any two tangents which are drawn from the same point to the circle is equal.

So,

(i) CF = CD = 6 cm

(ii) BE = BD = 8 cm

(iii) AE = AF = *x*

Now, it can be observed that,

(i) AB = EB+AE = 8+x

(ii) CA = CF+FA = 6+*x*

(iii) BC = DC+BD = 6+8 = 14

Now the semi perimeter โsโ will be calculated as follows

2s = AB+CA+BC

By putting the respective values we get,

2s = 28+2*x*

s = 14+*x*

By solving this we get,

= โ(14+*x*)48*x* โฆโฆโฆ (i)

Again, the area of โณABC = 2 ร area of (โณAOF + โณCOD + โณDOB)

= 2ร[(ยฝรOFรAF)+(ยฝรCDรOD)+(ยฝรDBรOD)]

= 2รยฝ(4*x*+24+32) = 56+4*x *โฆโฆโฆโฆ..(ii)

Now from (i) and (ii) we get,

โ(14+*x*)48*x *= 56+4*x*

Now, square both the sides,

48*x*(14+*x*) = (56+4*x*)^{2}

48*x = *[4(14+x)]^{2}/(14+*x*)

48*x = *16(14+*x*)

48*x = *224+16*x*

32*x = *224

*x = *7 cm

So, AB = 8+x

i.e. AB = 15 cm

And, CA = x+6 =13 cm.

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

**Answer:**

First draw a quadrilateral ABCD which will circumscribe a circle with its centre O in a way that it touches the circle at point P, Q, R, and S. Now, after joining the vertices of ABCD we get the following figure:

Now, consider the triangles OAP and OAS,

AP = AS (They are the tangents from the same point A)

OA = OA (It is the common side)

OP = OS (They are the radii of the circle)

So, by SSS congruency โณOAP โ โณOAS

So, โ POA = โ AOS

Which implies thatโ 1 = โ 8

Similarly, other angles will be,

โ 4 = โ 5

โ 2 = โ 3

โ 6 = โ 7

Now by adding these angles we get,

โ 1+โ 2+โ 3 +โ 4 +โ 5+โ 6+โ 7+โ 8 = 360ยฐ

Now by rearranging,

(โ 1+โ 8)+(โ 2+โ 3)+(โ 4+โ 5)+(โ 6+โ 7) = 360ยฐ

2โ 1+2โ 2+2โ 5+2โ 6 = 360ยฐ

Taking 2 as common and solving we get,

(โ 1+โ 2)+(โ 5+โ 6) = 180ยฐ

Thus, โ AOB+โ COD = 180ยฐ

Similarly, it can be proved that โ BOC+โ DOA = 180ยฐ

Therefore, the opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the center of the circle.

## NCERT Solutions for Class 10 Maths Chapter 10 Circles

This chapter comes under Unit 6 and has a weightage of 6 marks in the board examination. There will be one mark MCQ question, 2 marks reasoning questions and 3 marks short answer questions. This chapter has fundamental concepts that lay the foundation for your future studies.

The chapter Circle is included in Unit 4 Geometry of CBSE syllabus 2019-20. This unit has a weightage of 22 marks allotted in class 10 board examination. Unit 4 will have 4 MCQs carrying 4 marks, 2 Short answer questions carrying 6 marks and two long answer questions carrying 12 marks.

### Sub-topics of Class 10 Chapter 10 Circles

- Introduction to Circles
- Tangent to a circle
- Number of Tangents from a point on a circle
- Summary of the Whole Chapter

### List of Exercise from Class 10 Maths Chapter 10 Circles

Exercise 10.1โ 4 Questions which includes 1 short answer questions, 1 fill in the blanks question and 2 long answer questions

Exercise 10.2โ 13 Questions which includes 10 long answer questions, 4 descriptive type questions and 2 short answer questions

The Solutions for NCERT class 10 will guide you to understand the concepts involved in circles. You can refer this for better understanding of the concept. This Solution will also aid you to score good marks in the examination.

Class 10 Maths Chapter 10 deals with the existence of the tangents to a circle and some of the properties of circle. Students are introduced to some complex terms such as tangents, tangents to a circle, number of tangents from a point on the circle. This chapter seems very interesting due to the diagrams and involvement of geometrical calculations.

This NCERT Solution Class 10 Maths Chapter 10 has some tricky concept question on circles which will help you to clear all your doubts when you study. Students are recommended to study these solutions to know alternative calculation methods.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 10 Circles

- Provide answers to all the exercise questions in the NCERT class 10 Maths textbook.
- Create a practise of tricky questions which will clear your understanding of the topics.
- Covers entire syllabus and possible types of questions to be asked in the examination.
- Helps you to practise important drawings.
- Aids you in memorizing important formulas and calculation methods.

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## Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 10

### What are the key features of NCERT Solutions for Class 10 Maths Chapter 10?

NCERT Solutions for Class 10 Maths Chapter 10 provides solutions to all the exercise questions in the NCERT class 10 Maths Chapter 10. Also makes a practise of tricky questions which will clear your understanding of the topics.

### What are the main topics that are covered in NCERT Solutions for Class 10 Maths Chapter 10?

The main topics that are covered in NCERT Solutions for Class 10 Maths Chapter 10 are introduction to circles, tangent to a circle, number of tangents from a point on a circle and summary of the whole chapter.

### Is it necessary to learn all the questions provided in NCERT Solutions for Class 10 Maths Chapter 10?

Yes, you must learn all the questions provided in NCERT Solutions for Class 10 Maths Chapter 10. Because these questions may appear in board exams as well in class tests. By learning these questions students will be ready for their upcoming exams.