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Answers of Math NCERT solutions for Class 10th Maths Chapter 10 Circles

Unit 10

Circles 

Exercise 10.1, 10.2 Solutions

Exercise 10.1

Q1 :

How many tangents can a circle have?

Answer :

A circle can have infinite tangents.

Q2 :

Fill in the blanks:

  1.         A tangent to a circle intersects it in    points.
  2.         A line intersecting a circle in two points is called a    .
  3.     A circle can have parallel tangents at the most.
  4.     The common point of a tangent to a circle and the circle is called    .

Answer :

  1.         One
  2.     Secant
  3. Two
  4. Point of contact

Q3 :

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)

Answer :

We know that the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

OP ZAn9zoFJzs2g2AiPlu2IR4uAWVBZv0 lulVGLQ1gvYapaNrRQP15 6xHaHMLpC7u5aleoDct6W8ytKi 3X B laV6HKUxxgGfENXV6Fd50zpjS8CeqK uH7pBr10CXgbJIwSF1I PQtULHkVT55qSfm3lGoUcjNvEsV2mMiYsqK8snMGdX iJDEDTWGlAfqvCzDQ49QTzsnUagsRpMIzFq TjsCP1o9aBLXxxenwspy8ePfy629k3NnTa0Ch KCDbXuLbKauqDFBX4o5g

By applying Pythagoras theorem in ΔOPQ,Ww7QySL3vzX53hdgk05 xc4Pj6gdK5YLGr3hrzDwMfr K3a8U rjb83E5nNafQkM6bWP3l8N0Xr3yZRy88fn2EVK 4BgqqZylmBzWGDOPymX7clojX O2AizBb9hX9S2rYmBFGc

∴ OP2 + PQ2 = OQ2 52 + PQ2 =122

    PQ2 =144 – 25

     PQ =

     Hence, the correct answer is (D).

Q4 :

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 :kTmOrqmuEyLxkQ2O OpCj0 iOLYRKZK c w254

It can be observed that AB and CD are two parallel lines. Line AB is intersecting the circle at exactly two points, P and Q. Therefore, line AB is the secant of this circle. Since line CD is intersecting the circle at exactly one point, R, line CD is the tangent to the circle.

Exercise 10.2

Q1 :

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 :ej4pe06MqUtGomwF0yq 9it Vk5uANk1gHKi4wHE Yzg8P sGaToZo8K13s82T99V KUxfpbesV76drcHzcAxm8dzGeFFVrbFebqz9rDCi6YKr qaeVJBen20 gP2Jh8qkI Itg

Let O be the centre of the circle. Given that,

OQ = 25cm and PQ = 24 cm

As the radius is perpendicular to the tangent at the point of contact, Therefore, OP ⊥ PQ

Applying Pythagoras theorem in ΔOPQ, we obtain OP2 + PQ2 = OQ2

Bet SvwiaLBz613FrvpRb0psR725V0ViwZsr9vvX9KVXvRbroRcjvOn0LctvQNZDkdsUKsV1mIqvYV0X6shNLnl KVKKsZuHBG36gqaT yp822owo4fRQVWyku WeeR2 oF8u4
0ilPFZdaeCqJ kbUvHSHPmxwZCWvopPtKUk 0 UPjDX6Q OMGyKmhnD2pKRdoqmuqAX8T6b3hlaIYU1 LwTVbcrDuAiS0qE8GT KZkVxHNryTYgn OHKVIipLuE68Ce1YFgN8iM

OP = 7

Therefore, the radius of the circle is 7 cm. Hence, (A) is correct.

Q2 :

In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110 W7kifNe8WQ GhL4bgVie4c7N58NvsjZeGnhy9AhBsT sXGh3kJvW2VGaFxmlfkdj8qyzjZ1QU52lklu, then

∠PTQ is equal to 

(A) 60°              (B) 70°        (C) 80°          (D) 90°

Answer :

It is given that TP and TQ are tangents.

Therefore, radius drawn to these tangents will be perpendicular to the tangents. Thus, OP ⊥ TP and OQ ⊥ TQ

∠OPT = 90º

∠OQT = 90º

In quadrilateral POQT,

Sum of all interior angles = 360°

∠OPT + ∠POQ +∠OQT + ∠PTQ = 360°

⇒ 90º  + 110 º + 90º  + ∠PTQ = 360°

A JmZ65HKgRGOqauHmzgdS27xA96La jhJiGplMCXIxv83lq7mb96TCVUiYkBSXEf4o0JbqpxV nkEgnhOFazgvAlOMvr2HLehkgvTrwnKLY Me3wlnYwb2EEZ1FzupechvJtwPTQ = 70°

Hence, alternative (B) is correct.

Q3 :

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

(A) 50°            (B) 60°        (C) 70°         (D) 80°

Answer :

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents. Thus, OA ⊥ PA and OB ⊥ PB

∠OBP = 90 º

∠OAP = 90 º In AOBP,

Sum of all interior angles = 360°

∠OAP + ∠APB +∠PBO + ∠BOA = 360°

90°  + 80° + 90° + ∠BOA = 360°

∠BOA = 100°

In ΔOPB and ΔOPA,

AP = BP (Tangents from a point) 

OA = OB (Radii of the circle)

OP = OP (Common side)

Therefore, ΔOPB ~ ΔOPA (SSS congruence criterion) A ↔ B, P ↔ P, O ↔ O

And thus, ∠POB = ∠POA

NmSRXAy

Hence, alternative (A) is correct.

Q4 :

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

Answer :TP4ZWu1GfRkmIXoT7OT6hkrrj31n8eGnmp9woQPK7cYfFKKPYsoDRL 81XiXgRT7wVah6QZ2mcPM3LZ5D1KPPKDi t9a1wvHe6d9jzoW5xNFBhIB43MkMaTRFnQXKWS JsJrN3c

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively. Radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ RS and OB ⊥ PQ

∠ OAR = 90º

∠ OAS = 90º

∠ OBP = 90º

∠ OBQ = 90º

It can be observed that

∠ OAR = ∠ OBQ (Alternate interior angles)

∠ OAS = ∠ OBP (Alternate interior angles)

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

Q5 :

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

Answer:

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.TQB0ExMVuFDX2vC GsRJcwFz5ynih5f5tMy8ldoyME 2OE9W2 MLSoojyY9Jrrt eImuehThfhtxwAZ oh

We have to prove that the line perpendicular to AB at P passes through centre O. We shall prove this by contradiction method.

Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O’. Join OP and O’P.IeHO7obeCkwuFsqIPzbNF6SKhBGUNJS1Y44kXMaY1pzQFZb2VVIsx0Ip61yZgcddyCAcqM4fs A827CW0YpWt4d u H7K12AyeK3P5ninytCNRr8kfeAAH6IDpNuQkP4swGm5BA

As perpendicular to AB at P passes through O’, therefore,

∠ O’PB = 90°     … (1)

O is the centre of the circle and P is the point of contact. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.

∴ ∠ OPB = 90°     .. (2)

Comparing equations (1) and (2), we obtain

∠ O’PB = ∠ OPB     … (3)

From the figure, it can be observed that,

∠ O’PB < ∠ OPB     … (4)

Therefore, ∠ O’PB = ∠ OPB is not possible. It is only possible, when the line O’P coincides with OP. Therefore, the perpendicular to AB through P passes through centre O.

Q6 :

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 :5bUvKkgI7qimT7MEEFaRBREC3dvpeuv726DMNblDSN3WNO8zZ2DOs6gmx8xeVAk2vV GWa4IYS364a3iNXy 3NmC XufCANrvV 3NGCY3wDHjEJ3gzXab86ihHCvB5SAMt32XXM

Let us consider a circle centered at point O.

AB is a tangent drawn on this circle from point A. Given that,

OA = 5cm and AB = 4 cm In ΔABO,

OB ⊥ AB (Radius ⊥ tangent at the point of contact) Applying Pythagoras theorem in ΔABO, we obtain AB2 + BO2 = OA2

42 + BO= 52

16 + BO2 = 25

BO2 = 9

BO = 3

Hence, the radius of the circle is 3 cm.

*Q7 :

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 :

9EnUHvYu9tT5NrDkCyCpgac3 RBvYPGgSBYGdbAQlw950KT9egE0WhGZEwCYCMEBFl TzVaQBoi1knlTR5abUv2ZqcyfUuwVqJuAAH0ZzEs3GPzXp7SlPyEZqJWvLZ6SFaNreTA

Let the two concentric circles be centered at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to the smaller circle.

OA ⊥ PQ (As OA is the radius of the circle) Applying Pythagoras theorem in ΔOAP, we obtain OA2 + AP2 = OP2

32 + AP2 = 52

9 + AP2 = 25

AP2 = 16

AP = 4

In ΔOPQ,

Since OA ⊥ PQ,

AP = AQ (Perpendicular from the center of the circle bisects the chord) PQ = 2AP = 2 × 4 = 8

∴ Therefore, the length of the chord of the larger circle is 8 cm.

*Q8 :

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

Answer:

It can be observed that

DR = DS (Tangents on the circle from point D)         … (1)

CR = CQ (Tangents on the circle from point C)         … (2) 

BP = BQ (Tangents on the circle from point B)         … (3) 

AP = AS (Tangents on the circle from point A)         … (4)

 Adding all these equations, we obtain

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

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) CD + AB = AD + BCPZkFHyt9UmBh1nivS85T9lP ucMxNRUehgTel8fDF26il0Zdw1U4pt9GZHg79J9HHp378cgfYKT YtYkL97wmVftYY7KTsfJblUXrDoYFvC8BPDsWq4y7nvrTxt

*Q9 :

In the given figure, 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 W7kifNe8WQ GhL4bgVie4c7N58NvsjZeGnhy9AhBsT sXGh3kJvW2VGaFxmlfkdj8qyzjZ1QU52lklu.
DjxVFSDKlg45zAwfHg59hWkQefHrv6GThLMYKieTDuQx2TjGeacbYofGa4NrtQtfE4rJbx RiBk1eGBQ y7zAYz0jKz ekDgV UPy7AyPI5J6kOozzp pp6zGcXKXZrfGnHwuVI

Answer :

Let us join point O to C.
dRwatNVOS8Uj

In ΔOPA and ΔOCA,

OP = OC (Radii of the same circle)

 AP = AC (Tangents from point A) 

AO = AO (Common side)

ΔOPA ΔOCA (SSS congruence criterion)

Therefore, P ↔ C, A ↔ A, O ↔ O

∠POA = ∠COA         … (i) 

Similarly, ΔOQB ΔOCB

∠QOB = ∠COB         … (ii)

Since POQ is a diameter of the circle, it is a straight line. Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180 º From equations (i) and (ii), it can be observed that 2∠COA + 2 ∠COB = 180 º

∠COA + ∠COB = 90 º

∠AOB = 90°

Q10 :

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 :zfSbDdCY11qxTTXAxYH2wNeDqDVrT96o6ipMQX75AsQkQ hoh1uBlizIK2BgoY2UAkjTcMI3AI1rWrxGnhDj9 i8hnh58YhRTBOnd5EeJH3jUT5qgNJLD4 nzvNHJtwIQ56gsjs

Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠ AOB at center O of the circle.

It can be observed that

OA (radius) ⊥ PA (tangent) Therefore, ∠ OAP = 90°

Similarly, OB (radius) ⊥ PB (tangent)

∠ OBP = 90°

In quadrilateral OAPB,

Sum of all interior angles = 360º

∠ OAP +∠ APB+∠ PBO +∠ BOA = 360º 90º + ∠ APB + 90º + ∠ BOA = 360º

∠ APB + ∠ BOA = 180º

Hence, it can be observed 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.

Q11 :

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer :

Since ABCD is a parallelogram, AB = CD     …(1)

BC = AD                        …(2)fmcVTjRZKOTSOBBQZbxXTScEhj5LIiOvlYxv sOHITaXSpT6zc5sD34E1huq3QUhOIOVsaMI

It can be observed that

DR = DS (Tangents on the circle from point D) 

CR = CQ (Tangents on the circle from point C) 

BP = BQ (Tangents on the circle from point B) 

AP = AS (Tangents on the circle from point A)

 Adding all these equations, we obtain

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

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain 2AB = 2BC

AB = BC                         …(3)

Comparing equations (1), (2), and (3), we obtain AB = BC = CD = DA

Hence, ABCD is a rhombus.

*Q12 :

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 given figure). Find the sides AB and AC.

baFa4sbiGKa J KCSWdKjhWYBlfaZ YzIp9X0fFScPmVxNbEaLYrANO0uD9RLEyLCTha

Answer :

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be x.

In 3xtSSaCbJQjqFLk A91 Pmtmz2isZKJs0dqz91WVkWIQ39 4J Dgop3U66H1svpWLt 75 uEqoPSPRZbrX1VgRv1ec9Q8JP61eRUPGmDA 3k7NhEvV3WJTd0t3LVvsTDQN9W dwABC,

CF = CD = 6cm (Tangents on the circle from point C) 

BE = BD = 8cm (Tangents on the circle from point B) 

AE = AF = x (Tangents on the circle from point A) 

AB = AE + EB = x + 8

BC = BD + DC = 8 + 6 = 14 

CA = CF + FA = 6 + x

2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

s = 14 + x
FH1woxrUQ2AycYpq5Om47UBIZBxmbi3FRi7HH SWr4Iew9LTxoNhTeL nqNDIr1JUXnOboB5e8pi5equ6I2 UArGLZiCXsstf4F1rVTnNRZQf2hClYKfbsFu00ZTtm59zX0w3Qk

Area of

Area of

Area of

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOABEK73 PsJbMkqg5GaFpY49 sewLktHjZL1lgYtSiBc18BvPrXzW66ILho FWc1jjDZIV66rrP4Ti2i5kUVCZoC2HGlWteCnJtcv3Q IpPUQvadASzi8 GmC Oo4NoyURhv j3BG8

Either x+14 = 0 or x – 7 =0 Therefore, x = – 14and 7

However, x = – 14 is not possible as the length of the sides will be negative. Therefore, x = 7

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

CA = 6 + x = 6 + 7 = 13 cm

*Q13 :

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

Answer:

IZVVEgVMNeubDrux0gMJxoMyYQcQH nRnCS

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the center of the circle.

Consider ΔOAP and ΔOAS,

AP = AS (Tangents from the same point) OP = OS (Radii of the same circle)

OA = OA (Common side)

ΔOAP4nPRk41XLkun59Jkt 7tORSvOo LNi6v5cy87cwLjdHutwkT6tX07PsjJASWmqB AKYqQ2W7EqVUCQgf8VXE23azhouckqc6MMESVJ2nN7m6MooNUEVrNzΔOAS (SSS congruence criterion) 

Therefore, ΔOAP dv5Yxa S3AU6rWClxFoFK3RA9CExrP5F8e0diZ6zPJ81bCtMA2FjojKQM2BcOD13ChKs oTbtLkdGsMzhK EYv5h9TNgUfrcDtdmM9AyoGAeNiQhMGwVHAtheIFd266wiRSvah4 ΔOAS and thus, ∠ POA = ∠ AOS

∠ 1 = ∠ 8

Similarly,

∠ 2 = ∠ 3

∠ 4 = ∠ 5

∠ 6 = ∠ 7

∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360º

(∠ 1 + ∠ 8) + (∠ 2 + ∠ 3) + (∠ 4 + ∠ 5) + (∠ 6 + ∠ 7) = 360º

2∠ 1 + 2∠ 2 + 2∠ 5 + 2∠ 6 = 360º

2(∠ 1 + ∠ 2) + 2(∠ 5 + ∠ 6) = 360º

(∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180º

∠ AOB + ∠ COD = 180º

Similarly, we can prove that ∠ BOC + ∠ DOA = 180º

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre 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 10.1– 4 Questions which includes 1 short answer questions, 1 fill in the blanks question and 2 long answer questions

Exercise 10.2– 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.


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.


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NCERT Solutions Class 10 Maths Chapters

  • Chapter 1 Real Numbers
  • Chapter 2 Polynomials
  • Chapter 3 Pair Of Linear Equations In Two Variables
  • Chapter 4 Quadratic Equations
  • Chapter 5 Arithmetic Progressions
  • Chapter 6 Triangles
  • Chapter 7 Coordinate Geometry
  • Chapter 8 Introduction To Trigonometry
  • Chapter 9 Some Applications Of Trigonometry
  • Chapter 10 Circles
  • Chapter 11 Constructions
  • Chapter 12 Areas Related To Circles
  • Chapter 13 Surface Areas And Volumes
  • Chapter 14 Statistics
  • Chapter 15 Probability

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