# NCERT Solutions for Class 10 Maths Ch 10 Circles

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**NCERT Solutions for Class 10 Maths Ch 10 Circles**here which will be useful in preparing your answers in taking help. These NCERT Solutions for Class 10 are prepared buy Studyrankers experts who have large experience of teaching. These will be helpful in preparing yourself before examinations.
Page No: 209

**Exercise: 10.1**

1.Â How many tangents can a circle have?

**Answer**

A circle can have infinite tangents.

2. Â Fill in the blanks :

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

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

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

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

**Answer**

(i) one

(ii) secant

(iii) two

(iv) point of contact

3.Â 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 point Q so that OQ = 12 cm. Length PQ is :

(A) 12 cm

(B) 13 cm

(C) 8.5 cmÂ

(D) âˆš119Â cm

(B) 13 cm

(C) 8.5 cmÂ

(D) âˆš119Â cm

**Answer**

âˆ´ OP âŠ¥ PQ

By Pythagoras theorem inÂ Î”OPQ,

OQ

â‡’ (12)

^{2}= OP^{2}+^{Â }PQ^{2}â‡’ (12)

^{2Â }= 5^{2}+ PQ^{2}
â‡’PQ

^{2}= 144 - 25
â‡’PQ

^{2}Â =Â 119
â‡’PQÂ = âˆš119Â cm

(D) is the correct option.

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**

AB and XY are two parallel lines where AB is the tangent to the circle at point C while XY is the secant to the circle.

Page NO: 213

**Exercise: 10.2**

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

1. Â 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) Â 7 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

**Answer**

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

âˆ´ OP âŠ¥ PQ

also, Î”OPQ is right angled.

OQ = 25 cm and PQ = 24 cm (Given)

also, Î”OPQ is right angled.

OQ = 25 cm and PQ = 24 cm (Given)

By Pythagoras theorem inÂ Î”OPQ,

OQ

â‡’ (25)

^{2}Â = OP^{2}Â +^{Â }PQ^{2}â‡’ (25)

^{2Â }=Â OP^{2}Â + (24)^{2}
â‡’ OP

^{2}Â = 625 - 576
â‡’ OP

^{2}Â =Â 49
â‡’ OPÂ = 7 cm

The radius of the circle is option (A) 7 cm.

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

(A) 60Â°

(B) 70Â°Â

(C) 80Â°Â

(D) 90Â°

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

(A) 60Â°

(B) 70Â°Â

(C) 80Â°Â

(D) 90Â°

**Answer**

OP and OQ are radii of the circle to the tangents TP and TQ respectively.

âˆ´ OP âŠ¥ TP and,

âˆ´ OQ âŠ¥ TQ

âˆ OPT = âˆ OQT =Â 90Â°

In quadrilateral POQT,

Sum of all interior angles = 360Â°

âˆ PTQ +Â âˆ OPT + âˆ POQ + âˆ OQT Â = 360Â°

â‡’ âˆ PTQ +Â 90Â° + 110Â° + 90Â° Â = 360Â°

â‡’ âˆ PTQ = 70Â°

âˆ PTQ is equal toÂ option (B) 70Â°.

3. Â If tangents PA and PB from a point P to a circle with centre O are inclined to each otherÂ at angle of 80Â°, then âˆ POA is equal to

(A) 50Â° Â

(B) 60Â°Â

(C) 70Â°

(D) 80Â°

(A) 50Â° Â

(B) 60Â°Â

(C) 70Â°

(D) 80Â°

**Answer**

OA and OB are radii of the circle to the tangents PA and PB respectively.

âˆ´Â OAÂ âŠ¥Â PAÂ and,

âˆ´Â OBÂ âŠ¥ PB

âˆ OBP = âˆ OAP = 90Â°

âˆ´Â OAÂ âŠ¥Â PAÂ and,

âˆ´Â OBÂ âŠ¥ PB

âˆ OBP = âˆ OAP = 90Â°

In quadrilateral AOBP,

Sum of all interior angles = 360Â°

âˆ AOBÂ +Â âˆ OBPÂ +Â âˆ OAPÂ + âˆ APB Â = 360Â°

â‡’ âˆ AOBÂ +Â 90Â° + 90Â° + 80Â° Â = 360Â°

Sum of all interior angles = 360Â°

âˆ AOBÂ +Â âˆ OBPÂ +Â âˆ OAPÂ + âˆ APB Â = 360Â°

â‡’ âˆ AOBÂ +Â 90Â° + 90Â° + 80Â° Â = 360Â°

â‡’ âˆ AOBÂ = 100Â°

Now,

In Î”OPB and Î”OPA,AP = BP (Tangents from a point are equal)

OA = OB (Radii of the circle)

OP = OP (Common side)

âˆ´Â Î”OPB â‰… Î”OPA (by SSS congruence condition)

Thus âˆ POB = âˆ POA

âˆ AOB = âˆ POB +Â âˆ POA

â‡’ 2 âˆ POA = âˆ AOB

â‡’ âˆ POA = 100Â°/2 = 50Â°

âˆ POA is equal to option Â (A) 50Â°

Page No: 214

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

**Answer**

âˆ´ OB âŠ¥ RS and,

âˆ´ OA âŠ¥ PQ

âˆ OBR = âˆ OBS = âˆ OAP = âˆ OAQ = 90Âº

âˆ OBR = âˆ OBS = âˆ OAP = âˆ OAQ = 90Âº

From the figure,

âˆ OBR = âˆ OAQ (Alternate interior angles)

âˆ OBS = âˆ OAP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.

âˆ OBR = âˆ OAQ (Alternate interior angles)

âˆ OBS = âˆ OAP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.

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

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

**Answer**

Let AB be the tangent to the circle at point P with centre O.

We have to prove that PQ passes through the point O.

Suppose that PQ doesn't passes through point O. Join OP.

Through O, draw a straight line CD parallel to the tangent AB.

PQ intersect CD at R and also intersect AB at P.

AS, CD // AB PQ is the line of intersection,

âˆ ORP =Â âˆ RPA (Alternate interior angles)
but also,

âˆ RPA = 90Â° (PQÂ âŠ¥ AB)Â

â‡’ âˆ ORP Â = 90Â°

âˆ ROPÂ +Â âˆ OPA = 180Â° (Co-interior angles)

â‡’âˆ ROPÂ + 90Â° = 180Â°

â‡’âˆ ROP = 90Â°

Thus, the Î”ORP has 2 right angles i.e. âˆ ORP Â and âˆ ROP which is not possible.

Hence, our supposition is wrong.Â

âˆ´ PQ passes through the point O.

6. Â 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**

OA = 5cm and AB = 4 cm (Given)

In Î”ABO,

By Pythagoras theorem in Î”ABO,

OA

^{2}=

^{Â }AB

^{2Â }+ BO

^{2}

â‡’ 5

^{2Â }= 4

^{2Â }+ BO

^{2}

â‡’ BO

^{2}Â = 25 - 16

â‡’ BO

^{2}Â = 9

â‡’ BOÂ = 3

âˆ´ The radius of the circle 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**

AB be the chord of the larger circle which touches the smaller circle at point P.Â

âˆ´ AB is tangent to the smaller circle to the point P.

â‡’ OP âŠ¥ AB

By Pythagoras theorem in Î”OPA,

OA

â‡’ 5

â‡’Â AP

â‡’ AP = 4

In Î”OPB,

Since OP âŠ¥ AB,

AP = PB (Perpendicular from the center of the circle bisects the chord)

AB = 2AP = 2Â Ã— 4 = 8 cm

Â âˆ´ 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

From the figure we observe that,

DR = DS (Tangents on the circle from point D) â€¦ (i)

AP = AS (Tangents on the circle from point A) â€¦ (ii)

By Pythagoras theorem in Î”OPA,

OA

^{2}Â = Â AP^{2}Â + OP^{2}â‡’ 5

^{2}Â =Â AP^{2}Â + 3^{2}â‡’Â AP

^{2}Â = 25 - 9â‡’ AP = 4

In Î”OPB,

Since OP âŠ¥ AB,

AP = PB (Perpendicular from the center of the circle bisects the chord)

AB = 2AP = 2Â Ã— 4 = 8 cm

Â âˆ´ 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**

From the figure we observe that,

DR = DS (Tangents on the circle from point D) â€¦ (i)

AP = AS (Tangents on the circle from point A) â€¦ (ii)

BP = BQ (Tangents on the circle from point B) â€¦ (iii)

CR = CQ (Tangents on the circle from point C) â€¦ (iv)

Adding all these equations,

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

â‡’ (BP + AP) +Â (DR + CR) Â = (DS + AS) + (CQ + BQ)

CR = CQ (Tangents on the circle from point C) â€¦ (iv)

Adding all these equations,

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

â‡’ (BP + AP) +Â (DR + CR) Â = (DS + AS) + (CQ + BQ)

â‡’ CD + AB = AD + BC

9. In Fig. 10.13, XY and Xâ€²Yâ€² are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and Xâ€²Yâ€² at B. Prove that âˆ AOB = 90Â°.

**Answer**

We joined O and C

In Î”OPA and Î”OCA,

OP = OC (Radii of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

âˆ´ Î”OPAÂ â‰…Â Î”OCA (SSS congruence criterion)

â‡’ âˆ POA = âˆ COA â€¦ (i)

Similarly,

Â Î”OQBÂ Â â‰…Â Î”OCB

âˆ QOB = âˆ COB â€¦ (ii)

Since POQ is a diameter of the circle, it is a straight line.

âˆ´ âˆ POA + âˆ COA + âˆ COB + âˆ QOB = 180 Âº

From equations (i) and (ii),

2âˆ COA + 2âˆ COB = 180Âº

â‡’ âˆ COA + âˆ COB = 90Âº

â‡’ âˆ AOB = 90Â°

âˆ QOB = âˆ COB â€¦ (ii)

Since POQ is a diameter of the circle, it is a straight line.

âˆ´ âˆ POA + âˆ COA + âˆ COB + âˆ QOB = 180 Âº

From equations (i) and (ii),

2âˆ COA + 2âˆ COB = 180Âº

â‡’ âˆ COA + âˆ COB = 90Âº

â‡’ âˆ AOB = 90Â°

10. Â 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.

**Answer**

It can be observed that

OA âŠ¥ PA

âˆ´ âˆ OAP = 90Â°

Similarly, OB âŠ¥ PB

âˆ´ âˆ OBP = 90Â°

In quadrilateral OAPB,

Sum of all interior angles = 360Âº

âˆ OAP +âˆ APB +âˆ PBO +âˆ BOA = 360Âº

â‡’ 90Âº + âˆ APB + 90Âº + âˆ BOA = 360Âº

â‡’ âˆ APB + âˆ BOA = 180Âº

âˆ´ 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.

ABCD is a parallelogram,

âˆ´ AB = CD ... (i)

âˆ´ BC = AD ... (ii)

CR = CQ (Tangents to the circle atÂ C)

BP = BQ (Tangents to the circle atÂ B)

AP = AS (Tangents to the circle atÂ A)

Adding all these,

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

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

InÂ Î”ABC,

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

**Answer**

âˆ´ AB = CD ... (i)

âˆ´ BC = AD ... (ii)

From the figure, we observe that,

DR = DS (Tangents to the circle at D)CR = CQ (Tangents to the circle atÂ C)

BP = BQ (Tangents to the circle atÂ B)

AP = AS (Tangents to the circle atÂ A)

Adding all these,

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

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

â‡’ CD + AB = AD + BC ... (iii)

Putting the value of (i) and (ii) in equation (iii) we get,

2AB = 2BC

â‡’ AB = BC ... (iv)

By Comparing equations (i), (ii), and (iv) we get,

AB = BC = CD = DA

âˆ´Â ABCD is a rhombus.

Putting the value of (i) and (ii) in equation (iii) we get,

2AB = 2BC

â‡’ AB = BC ... (iv)

By Comparing equations (i), (ii), and (iv) we get,

AB = BC = CD = DA

âˆ´Â ABCD is a rhombus.

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD andÂ 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**

Length of two tangents drawn from the same point to the circle are equal,

âˆ´ CF = CD = 6cm

âˆ´ BE = BD = 8cm

âˆ´ AE = AF =

=Â âˆš(14 + âˆ´ CF = CD = 6cm

âˆ´ BE = BD = 8cm

âˆ´ AE = AF =

*x*
We observed that,

AB = AE + EB =

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 +

AB = AE + EB =

*x*+ 8BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 +

*x*
Now semi perimeter of triangle (s) is,

â‡’ 2s = AB + BC + CA

=

= 28 + 2

â‡’s = 14 +Â

â‡’ 2s = AB + BC + CA

=

*x*+ 8 + 14 + 6 +*x*= 28 + 2

*x*â‡’s = 14 +Â

*x*
Area of Î”ABC = âˆšs (s - a)(s - b)(s - c)

*x*)Â (14 +Â

*x*-

*14)(14 +Â*

*xÂ*-

*Â x*- 6)(14 +Â

*xÂ*-

*Â x -Â*8)

= âˆš(14 +Â

= âˆš(14 +Â *x*)Â (*x*)(8)(6)*x*) 48Â

*x*Â ... (i)

also, Area of Î”ABC = 2Ã—area of (Î”AOFÂ +Â Î”CODÂ +Â Î”DOB)

= 2Ã—[(1/2Ã—OFÃ—AF)Â + (1/2Ã—CDÃ—OD)Â + (1/2Ã—DBÃ—OD)]

= 2Ã—1/2 (4

*x*+ 24Â + 32) = 56 + 4*xÂ*... (ii)
Equating equation (i) and (ii) we get,

âˆš(14 +Â

Squaring both sides,*x*) 48Â*xÂ*= 56 + 4*x*
48

*x*(14 +*x*) = (56 + 4*x*)^{2}
â‡’ 48

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

*x =Â*16 (14 +Â*x*)
â‡’ 48

*x =Â*224 + 16*x*
â‡’ 32

*x =Â*224
â‡’

Hence, AB = *x =Â*7 cm*x*+ 8 = 7 + 8 = 15 cm

CA = 6 +

*x*= 6 + 7 = 13 cm

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

**Answer**

In Î”OAP and Î”OAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the circle)

OA = OA (Common side)

Î”OAP â‰… Î”OAS (SSS congruence condition)

âˆ´Â âˆ POA = âˆ AOS

â‡’âˆ 1 = âˆ 8

Similarly we get,

âˆ 2 = âˆ 3

âˆ 4 = âˆ 5

âˆ 6 = âˆ 7

Similarly we get,

âˆ 2 = âˆ 3

âˆ 4 = âˆ 5

âˆ 6 = âˆ 7

Adding all these angles,

âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 + âˆ 5 + âˆ 6 + âˆ 7 +âˆ 8 = 360Âº

â‡’ (âˆ 1 + âˆ 8) + (âˆ 2 + âˆ 3) + (âˆ 4 + âˆ 5) + (âˆ 6 + âˆ 7) = 360Âº

â‡’ 2 âˆ 1 + 2 âˆ 2 + 2 âˆ 5 + 2 âˆ 6 = 360Âº

â‡’ 2(âˆ 1 + âˆ 2) + 2(âˆ 5 + âˆ 6) = 360Âº

â‡’ (âˆ 1 + âˆ 2) + (âˆ 5 + âˆ 6) = 180Âº

â‡’ âˆ AOB + âˆ COD = 180Âº

Similarly, we can prove that âˆ BOC + âˆ DOA = 180Âº

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 + âˆ 5 + âˆ 6 + âˆ 7 +âˆ 8 = 360Âº

â‡’ (âˆ 1 + âˆ 8) + (âˆ 2 + âˆ 3) + (âˆ 4 + âˆ 5) + (âˆ 6 + âˆ 7) = 360Âº

â‡’ 2 âˆ 1 + 2 âˆ 2 + 2 âˆ 5 + 2 âˆ 6 = 360Âº

â‡’ 2(âˆ 1 + âˆ 2) + 2(âˆ 5 + âˆ 6) = 360Âº

â‡’ (âˆ 1 + âˆ 2) + (âˆ 5 + âˆ 6) = 180Âº

â‡’ âˆ AOB + âˆ COD = 180Âº

Similarly, we can prove that âˆ BOC + âˆ DOA = 180Âº

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

**Go Back To NCERT Solutions for Class 10th Math**### NCERT Solutions for Class 10 Maths Chapter 10 Circles

Chapter 10 Class 10 Maths is very important for the purpose of examinations. There are total 3 topics in this chapter through you will learn more about circles.

â€¢ Introduction: A circle is a collection of all points in a plane which are at a constant distance (radius) from a fixed point (centre). You will learn about various key terms such asÂ Â related to a circle like chord, segment, sector, arc etc.

â€¢Â Tangent to a Circle: A tangent to a circle is a line that intersects the circle at only one point. There is

only one tangent at a point of the circle. The are various problems related to this topics that includes fill in the blanks and finding the length of a given line.

â€¢Â Number of Tangents from a Point on a Circle: There is one theorem in this topics that states the lengths of tangents drawn from an external point to a circle are equal. The questions includes MCQs and related proving and finding.

If you want to access other study resources prepared by Studyrankers then you can follow the provided links.

CBSE Notes for Class 10 Maths Chapter 10 Circles

MCQ Questions for Classs 10 Maths Chapter 10 Circles

There are total 2 exercises in the chapter which will check you basic understanding about the circles. These NCERT Solutions for Class 10 are best way to encourage students to learn new topics. You can also find exercisewise solutions just by clicking on the links given below.

- Exercise 10.1
- Exercise 10.2

**NCERT Solutions for Class 10 Maths Chapters:**

**FAQ onÂ Chapter 10 Circles**

#### How many exercises in Chapter 10 Circles

There are only 2 exercise in the Chapter 10 Class 10 Maths which will helpful in completing the homework on time and increasing concentration among students.

#### What isÂ point of contact?

A line which touches a circle only in one point is called a tangent line and the point at which it touches the circle is called the point of contact.

#### The ______ at any point of a circle is perpendicular to the radius, through the point of contact.

The tangent at any point of a circle is perpendicular to the radius, through the point of contact.

#### What do you mean by Tangent?

A tangent to a circle is a line that touches the circle at only one point.