NCERT Solutions Class 10 Mathematics Chapter 10 Circles

NCERT Solutions Class 10 Mathematics Chapter 10 Circles have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 10 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 10 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 10 Mathematics are an important part of exams for Class 10 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 10 Mathematics and also download more latest study material for all subjects. Chapter 10 Circles is an important topic in Class 10, please refer to answers provided below to help you score better in exams

Chapter 10 Circles Class 10 Mathematics NCERT Solutions

Class 10 Mathematics students should refer to the following NCERT questions with answers for Chapter 10 Circles in Class 10. These NCERT Solutions with answers for Class 10 Mathematics will come in exams and help you to score good marks

Chapter 10 Circles NCERT Solutions Class 10 Mathematics

Exercise 10.1

Q.1) How many tangents can a circle have?
Sol.1) A circle can have infinite tangents.

Q.2) 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 ............
Sol.1) (i) one (ii) secant (iii) two (iv) point of contact

Q.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 π‘π‘š. Length PQ is :
(A) 12 cm (B) 13 cm (C) 8.5 cm (D) √119 π‘π‘š
Sol.3) The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ 𝑂𝑃 βŠ₯ 𝑃𝑄
By Pythagoras theorem in π›₯𝑂𝑃𝑄,
𝑂𝑄2 = 𝑂𝑃2 + 𝑃𝑄2
β‡’ (12)2 = 52 + 𝑃𝑄2
β‡’ 𝑃𝑄2 = 144 βˆ’ 25
β‡’ 𝑃𝑄2 = 119
β‡’ 𝑃𝑄 = √119 π‘π‘š
(D) is the correct option

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles

Q.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.
Sol.4) 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

Exercise: 10.2

Q.1) In Q.1 to 3, choose the correct option and give justification.
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
Sol.1) The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ 𝑂𝑃 βŠ₯ 𝑃𝑄
also, π›₯𝑂𝑃𝑄 is right angled.
𝑂𝑄 = 25 π‘π‘š and 𝑃𝑄 = 24 π‘π‘š (Given)
By Pythagoras theorem in π›₯𝑂𝑃𝑄,
𝑂𝑄2 = 𝑂𝑃2 + 𝑃𝑄2
β‡’ (25)2 = 𝑂𝑃2 + (24)2
β‡’ 𝑂𝑃2 = 625 βˆ’ 576
β‡’ 𝑂𝑃2 = 49
β‡’ 𝑂𝑃 = 7 π‘π‘š
The radius of the circle is option (A) 7 cm.

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-1

Q.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°
Sol.2) OP and OQ are radii of the circle to the tangents TP and TQ respectively.
∴ 𝑂𝑃 βŠ₯ 𝑇𝑃 and,
∴ 𝑂𝑄 βŠ₯ 𝑇𝑄 βˆ π‘‚π‘ƒπ‘‡ = βˆ π‘‚π‘„π‘‡ = 90Β°
In quadrilateral 𝑃𝑂𝑄𝑇, Sum of all interior angles = 360Β°
βˆ π‘ƒπ‘‡π‘„ + βˆ π‘‚π‘ƒπ‘‡ + βˆ π‘ƒπ‘‚π‘„ + βˆ π‘‚π‘„π‘‡ = 360Β°
β‡’ βˆ π‘ƒπ‘‡π‘„ + 90Β° + 110Β° + 90Β° = 360Β°
β‡’ βˆ π‘ƒπ‘‡π‘„ = 70Β°
βˆ π‘ƒπ‘‡π‘„ is equal to option (B) 70Β°.

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-2

Q.3) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80Β°, then ∠ 𝑃𝑂𝐴 is equal to
(A) 50Β° (B) 60Β° (C) 70Β° (D) 80Β°
Sol.3) OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ 𝑂𝐴 βŠ₯ 𝑃𝐴 and,
∴ 𝑂𝐡 βŠ₯ 𝑃𝐡 βˆ π‘‚π΅π‘ƒ = βˆ π‘‚π΄π‘ƒ = 90Β°
In quadrilateral AOBP, Sum of all interior angles = 360Β°
βˆ π΄π‘‚π΅ + βˆ π‘‚π΅π‘ƒ + βˆ π‘‚π΄π‘ƒ + βˆ π΄π‘ƒπ΅ = 360Β°
β‡’ βˆ π΄π‘‚π΅ + 90Β° + 90Β° + 80Β° = 360Β°
β‡’ βˆ π΄π‘‚π΅ = 100Β°
Now, In π›₯𝑂𝑃𝐡 and π›₯𝑂𝑃𝐴,
𝐴𝑃 = 𝐡𝑃               (Tangents from a point are equal)
𝑂𝐴 = 𝑂𝐡               (Radii of the circle)
𝑂𝑃 = 𝑂𝑃               (Common side)
∴ π›₯𝑂𝑃𝐡 β‰… π›₯𝑂𝑃𝐴     (by SSS congruence condition)
Thus βˆ π‘ƒπ‘‚π΅ = βˆ π‘ƒπ‘‚π΄ βˆ π΄π‘‚π΅ = βˆ π‘ƒπ‘‚π΅ + βˆ π‘ƒπ‘‚π΄
β‡’ 2 βˆ π‘ƒπ‘‚π΄ = βˆ π΄π‘‚π΅
β‡’ βˆ π‘ƒπ‘‚π΄ = 100Β°/2 = 50Β°
βˆ π‘ƒπ‘‚π΄ is equal to option (A) 50Β°

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-3

Q.4) Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Sol.4) Let AB be a diameter of the circle.
Two tangents PQ and RS are drawn at points A and B respectively.
Radii of the circle to the tangents will be perpendicular to it.
∴ 𝑂𝐡 βŠ₯ 𝑅𝑆 and,
∴ 𝑂𝐴 βŠ₯ 𝑃𝑄
βˆ π‘‚π΅π‘… = βˆ π‘‚π΅π‘† = βˆ π‘‚π΄π‘ƒ = βˆ π‘‚π΄π‘„ = 90Β°
From the figure, βˆ π‘‚π΅π‘… = βˆ π‘‚π΄π‘„      (Alternate interior angles)
βˆ π‘‚π΅π‘† = βˆ π‘‚π΄π‘ƒ                                 (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.

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-4

Q.5) Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Sol.5) 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, 𝐢𝐷 // 𝐴𝐡
𝑃𝑄 is the line of intersection,
βˆ π‘‚π‘…π‘ƒ = βˆ π‘…π‘ƒπ΄ (Alternate interior angles) but also,
βˆ π‘…π‘ƒπ΄ = 90Β° (𝑃𝑄 βŠ₯ 𝐴𝐡)
β‡’ βˆ π‘‚π‘…π‘ƒ = 90Β° βˆ π‘…π‘‚π‘ƒ + βˆ π‘‚π‘ƒπ΄ = 180Β° (Co-interior angles)
β‡’ βˆ π‘…π‘‚π‘ƒ + 90Β° = 180Β°
β‡’ βˆ π‘…π‘‚π‘ƒ = 90Β°
Thus, the π›₯𝑂𝑅𝑃 has 2 right angles i.e.
βˆ π‘‚π‘…π‘ƒ and βˆ π‘…π‘‚π‘ƒ which is not possible.
Hence, our supposition is wrong.
∴ 𝑃𝑄 passes through the point O.

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-5

Q.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.
Sol.6) AB is a tangent drawn on this circle from point 𝐴.
∴ 𝑂𝐡 βŠ₯ 𝐴𝐡
𝑂𝐴 = 5π‘π‘š and 𝐴𝐡 = 4 π‘π‘š (Given)
In π›₯𝐴𝐡𝑂, By Pythagoras theorem in
π›₯𝐴𝐡𝑂, 𝑂𝐴2 = 𝐴𝐡2 + 𝐡𝑂2
β‡’ 52 = 42 + 𝐡𝑂2
β‡’ 𝐡𝑂2 = 25 βˆ’ 16
β‡’ 𝐡𝑂2 = 9
β‡’ 𝐡𝑂 = 3
∴ The radius of the circle is 3 cm

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-6

Q.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
Sol.7) Let the two concentric circles with centre 𝑂.
𝐴𝐡 be the chord of the larger circle which touches the smaller circle at point P.
∴ 𝐴𝐡 is tangent to the smaller circle to the point 𝑃.
β‡’ 𝑂𝑃 βŠ₯ 𝐴𝐡
By Pythagoras theorem in π›₯𝑂𝑃𝐴,
𝑂𝐴2 = 𝐴𝑃2 + 𝑂𝑃2
β‡’ 52 = 𝐴𝑃2 + 32
β‡’ 𝐴𝑃2 = 25 βˆ’ 9
β‡’ 𝐴𝑃 = 4
In π›₯𝑂𝑃𝐡, Since 𝑂𝑃 βŠ₯ 𝐴𝐡,
𝐴𝑃 = 𝑃𝐡 (Perpendicular from the center of the circle bisects the chord)
𝐴𝐡 = 2𝐴𝑃 = 2 Γ— 4 = 8 π‘π‘š
∴ The length of the chord of the larger circle is 8 cm

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-7

Q.8) A quadrilateral ABCD is drawn to circumscribe a circle (see Fig.). Prove that 𝐴𝐡 + πΆπ· = 𝐴𝐷 + 𝐡𝐢
Sol.8) From the figure we observe that,
DR = DS (Tangents on the circle from point D)        … (i)
AP = AS (Tangents on the circle from point A)         β€¦ (ii)
BP = BQ (Tangents on the circle from point B)        … (iii)
CR = CQ (Tangents on the circle from point C)       … (iv)
Adding all these equations,
𝐷𝑅 + 𝐴𝑃 + 𝐡𝑃 + 𝐢𝑅 = 𝐷𝑆 + 𝐴𝑆 + 𝐡𝑄 + 𝐢𝑄
β‡’ (𝐡𝑃 + 𝐴𝑃) + (𝐷𝑅 + 𝐢𝑅) = (𝐷𝑆 + 𝐴𝑆) + (𝐢𝑄 + 𝐡𝑄)
β‡’ 𝐢𝐷 + 𝐴𝐡 = 𝐴𝐷 + 𝐡𝐢

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-8

Q.9) In the given figure, π‘‹π‘Œ and π‘‹β€™π‘Œβ€™ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting π‘‹π‘Œ at A and π‘‹β€™π‘Œβ€™ at B.
Sol.9) We joined O and C
A/q,
In π›₯𝑂𝑃𝐴 and π›₯𝑂𝐢𝐴,
𝑂𝑃 = 𝑂𝐢         (Radii of the same circle)
𝐴𝑃 = 𝐴𝐢         (Tangents from point A)
𝐴𝑂 = 𝐴𝑂         (Common side)
∴ π›₯𝑂𝑃𝐴 β‰… π›₯𝑂𝐢𝐴 (SSS congruence criterion)
β‡’ βˆ π‘ƒπ‘‚π΄ = βˆ πΆπ‘‚π΄ … (i)
Similarly, π›₯𝑂𝑄𝐡 β‰… π›₯𝑂𝐢𝐡
βˆ π‘„π‘‚π΅ = βˆ πΆπ‘‚π΅ … (ii)
Since POQ is a diameter of the circle, it is a straight line.
∴ βˆ π‘ƒπ‘‚π΄ + βˆ πΆπ‘‚π΄ + βˆ πΆπ‘‚π΅ + βˆ π‘„π‘‚π΅ = 180Β°
From equations (i) and (ii),
2βˆ πΆπ‘‚π΄ + 2βˆ πΆπ‘‚π΅ = 180Β°
β‡’ βˆ πΆπ‘‚π΄ + βˆ πΆπ‘‚π΅ = 90Β°
β‡’ βˆ π΄π‘‚π΅ = 90Β°

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-9

Q.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.
Sol.10) 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 βˆ π΄π‘‚π΅ at center O of the circle.
It can be observed that
𝑂𝐴 (radius) βŠ₯ 𝑃𝐴      (tangent)
Therefore, βˆ π‘‚π΄π‘ƒ = 90Β°
Similarly, 𝑂𝐡           (radius) βŠ₯ 𝑃𝐡 (tangent)
βˆ π‘‚π΅π‘ƒ = 90Β°
In quadrilateral 𝑂𝐴𝑃𝐡,
Sum of all interior angles = 360Β°
βˆ π‘‚π΄π‘ƒ + βˆ π΄π‘ƒπ΅ + βˆ π‘ƒπ΅π‘‚ + βˆ π΅π‘‚π΄ = 360Β°
90Β° + βˆ π΄π‘ƒπ΅ + 180Β° + βˆ π΅π‘‚π΄ = 360Β°

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-10

βˆ π΄π‘ƒπ΅ + βˆ π΅π‘‚π΄ = 180Β°
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.

Q.11) Prove that the parallelogram circumscribing a circle is a rhombus.
Sol.11) ABCD is a parallelogram,
∴ 𝐴𝐡 = 𝐢𝐷 ... (i)
∴ 𝐡𝐢 = 𝐴𝐷 ... (ii)
From the figure, we observe that,
DR = DS (Tangents to the circle at D)
CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A) Adding all these,
𝐷𝑅 + 𝐢𝑅 + 𝐡𝑃 + 𝐴𝑃 = 𝐷𝑆 + 𝐢𝑄 + 𝐡𝑄 + 𝐴𝑆
β‡’ (𝐷𝑅 + 𝐢𝑅) + (𝐡𝑃 + 𝐴𝑃) = (𝐷𝑆 + 𝐴𝑆) + (𝐢𝑄 + 𝐡𝑄)
β‡’ 𝐢𝐷 + 𝐴𝐡 = 𝐴𝐷 + 𝐡𝐢 ... (iii)
Putting the value of (i) and (ii) in equation (iii) we get,
2𝐴𝐡 = 2𝐡𝐢
β‡’ 𝐴𝐡 = 𝐡𝐢 ... (iv)
By Comparing equations (i), (ii), and (iv) we get,
𝐴𝐡 = 𝐡𝐢 = 𝐢𝐷 = 𝐷𝐴
∴ 𝐴𝐡𝐢𝐷 is a rhombus.

Q.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.). Find the sides AB and AC.
Sol.12) In π›₯𝐴𝐡𝐢,
Length of two tangents drawn from the same point to the circle are equal,
∴ 𝐢𝐹 = 𝐢𝐷 = 6π‘π‘š
∴ 𝐡𝐸 = 𝐡𝐷 = 8π‘π‘š
∴ 𝐴𝐸 = 𝐴𝐹 = π‘₯
We observed that,
𝐴𝐡 = 𝐴𝐸 + 𝐸𝐡 = π‘₯ + 8
𝐡𝐢 = 𝐡𝐷 + 𝐷𝐢 = 8 + 6 = 14
𝐢𝐴 = 𝐢𝐹 + 𝐹𝐴 = 6 + π‘₯
Now semi perimeter of circles,
β‡’ 2𝑠 = 𝐴𝐡 + 𝐡𝐢 + 𝐢𝐴
= π‘₯ + 8 + 14 + 6 + π‘₯ = 28 + 2π‘₯
β‡’ 𝑠 = 14 + π‘₯

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-12

= 2 Γ— 1/2
(4π‘₯ + 24 + 32) = 56 + 4π‘₯ ... (ii)
Equating equation (i) and (ii) we get,
√(14 + π‘₯)48π‘₯ = 56 + 4π‘₯
Squaring both sides,
48π‘₯ (14 + π‘₯) = (56 + 4π‘₯)2
β‡’ 48π‘₯ = [4(14 + π‘₯)]2/14 + π‘₯
β‡’ 48π‘₯ = 16 (14 + π‘₯)
β‡’ 48π‘₯ = 224 + 16π‘₯
β‡’ 32π‘₯ = 224
β‡’ π‘₯ = 7 π‘π‘š
Hence, 𝐴𝐡 = π‘₯ + 8 = 7 + 8 = 15 π‘π‘š
𝐢𝐴 = 6 + π‘₯ = 6 + 7 = 13 π‘π‘š

Q.13) Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Sol.13) Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle
at point P, Q, R, S.
Join the vertices of the quadrilateral ABCD to the center of the circle.
In π›₯𝑂𝐴𝑃 and π›₯𝑂𝐴𝑆,
𝐴𝑃 = 𝐴𝑆 (Tangents from the same point)
𝑂𝑃 = 𝑂𝑆 (Radii of the circle)
𝑂𝐴 = 𝑂𝐴 (Common side)
π›₯𝑂𝐴𝑃 β‰… π›₯𝑂𝐴𝑆 (SSS congruence condition)
∴ βˆ π‘ƒπ‘‚π΄ = βˆ π΄π‘‚π‘†
β‡’ ∠1 = ∠8
Similarly, we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
Adding all these angles,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
β‡’ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360Β°
β‡’ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360Β°
β‡’ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360Β°
β‡’ (∠1 + ∠2) + (∠5 + ∠6) = 180Β°
β‡’ βˆ π΄π‘‚π΅ + βˆ πΆπ‘‚π· = 180Β°
Similarly, we can prove that ∠ 𝐡𝑂𝐢 + ∠ 𝐷𝑂𝐴 = 180Β°
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle

""NCERT-Solutions-Class-10-Mathematics-Chapter-10-Circles-13

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Chapter 10 Circles Class 10 NCERT Solution Mathematics

These solutions of Chapter 10 Circles NCERT Questions given in your textbook for Class 10 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 10.

Class 10 NCERT Solution Mathematics Chapter 10 Circles

NCERT Solutions Class 10 Mathematics Chapter 10 Circles detailed answers are given with the objective of helping students compare their answers with the example. NCERT solutions for Class 10 Mathematics provide a strong foundation for every chapter. They ensure a smooth and easy knowledge of Revision notes for Class 10 Mathematics. As suggested by the HRD ministry, they will perform a major role in JEE. Students can easily download these solutions and use them to prepare for upcoming exams and also go through the Question Papers for Class 10 Mathematics to clarify all doubts

Where can I download latest NCERT Solutions for Class 10 Mathematics Chapter 10 Circles

You can download the NCERT Solutions for Class 10 Mathematics Chapter 10 Circles for latest session from StudiesToday.com

Can I download the NCERT Solutions of Class 10 Mathematics Chapter 10 Circles in Pdf

Yes, you can click on the link above and download NCERT Solutions in PDFs for Class 10 for Mathematics Chapter 10 Circles

Are the Class 10 Mathematics Chapter 10 Circles NCERT Solutions available for the latest session

Yes, the NCERT Solutions issued for Class 10 Mathematics Chapter 10 Circles have been made available here for latest academic session

How can I download the Chapter 10 Circles Class 10 Mathematics NCERT Solutions

You can easily access the links above and download the Chapter 10 Circles Class 10 NCERT Solutions Mathematics for each chapter

Is there any charge for the NCERT Solutions for Class 10 Mathematics Chapter 10 Circles

There is no charge for the NCERT Solutions for Class 10 Mathematics Chapter 10 Circles you can download everything free

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Regular revision of NCERT Solutions given on studiestoday for Class 10 subject Mathematics Chapter 10 Circles can help you to score better marks in exams

Are there any websites that offer free NCERT solutions for Chapter 10 Circles Class 10 Mathematics

Yes, studiestoday.com provides all latest NCERT Chapter 10 Circles Class 10 Mathematics solutions based on the latest books for the current academic session

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Are NCERT solutions for Class 10 Chapter 10 Circles Mathematics available in multiple languages

Yes, NCERT solutions for Class 10 Chapter 10 Circles Mathematics are available in multiple languages, including English, Hindi