NCERT Solutions for Class 10 Maths Chapter 10 Circles

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Updated at 2021-02-23


NCERT solutions for class 10 Maths Chapter 10 Circles Exercise - 10.1

Q1 ) How many tangents can a circle have ?



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :

A circle can have an infinite number of tangents.

Q2 ) 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 ______.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :


(i) exactly one

(ii) secant

(iii) two

(iv) 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 of PQ is :
(A) 12 cm

(B) 13 cm

(C) 8.5 cm

(D) 119 cm



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :




PQ = OQ2 ? OP2 
= 122 ? 52 
= 144 ? 25  
= 119cm

So, (D) option is correct

Q4 ) Draw a circle and two lines parallel to a given line such that one is tangent and the other, a secant to the circle.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :



NCERT solutions for class 10 Maths Chapter 10 Circles 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



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :




Since QT is a tangent to the circle at T and OT is radius,

? OT?QT

It is given that OQ = 25 cm and QT = 24 cm,

? By Pythagoras theorem, we have

?OQ2 = QT2 + OT2
?OT2 = OQ2 ? QT2
? OT2 = 252 ? 242
?OT2 = 49

? OT = 49 = 7

? radius of the circle is 7 cm

? Option (A) is correct.

Q2 ) In figure, 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°



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :




Since TP and TQ are tangents to a circle with centre O so that ? POQ = 110°,

?OP ? PT and OQ ? QT

? ? OPT = 90° and ? OQT = 90°

In the quadrilateral TPOQ, we have

? PTQ + ? TPO + ? POQ + ? OQT = 360°
[ \therefore Sum of all angles in a quadrilateral = 360° ]

? PTQ + 90° + 110° + 90° = 360°

? PTQ + 290° = 360°

? ? PTQ = 360° ? 290°

? ? PTQ = 70°

? Option (B) is correct.

Q3 ) 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°



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :




Since PA and PB are tangents to a circle with centre O,

?OA ? AP, and OB ? BP

? OAP = 90° and ? OBP=90°

In the quadrilateral PAOB, we have

?? ABP + ? PAO + ? AOB + ? OBP = 360°

? ? 80° + 90° + ? AOB + 90° = 360°

? 260° + ? AOB = 360°

? ? AOB = 360° ? 260° = 100°

In ? OAP and OBP, we have

OP = OP [Common]

OA = OB [Radii]

? OAP = ? OBP [Each = 90°]

? ? OAP ? ? OBP ( By SAS Criterion)

? ? POA = ? POB [C.P.C.T.]

? ? POA = 12 ? AOB 
= 12 × 100° = 50°

? The (A) is the correct option.

Q4 ) Prove that the tangents drawn at the ends of a diameter of a circle are parallel.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :




Let PQ be a diameter of the given circle with centre O.

Let AB and CD be the tangents drawn to the circle at the end points of the diameter PQ respectively.

? Tangent at a point to a circle is perpendicular to the radius through the point,

? PQ ? AB and PQ ? CD

? ? APQ = ? PQD

? AB || CD [? ? APQ and ? PQD are alternate angles]

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

Q5 ) Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :




Let AB be the tangent drawn at the point P on the circle with O.

If possible, let PQ be perpendicular to AB, not passing through O. Join OP.

? Tangent at a point to a circle is perpendicular to the radius through the point,

? AB ? OP
?? OPB = 90°

? QPB = 90° (Construction)

?? QPB = ? OPB , which is not possible.

? It contradicts our supposition.

Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

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.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :


Since tangent to a circle is perpendicular to the radius through the point of contact,

? ? OTA = 90°

In ? OTA, we have

?OA2 = OT2 + AT2 
? 52 = OT2 + 42
? OT2 = 52 ? 42 = 25 ? 16 = 9

OT = 9 = 3

? 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.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :


Let O be the common centre of two concentric circles, and let AB be a chord of the larger circle touching the smaller circle at P. Join OP.



? OP is the radius of the smaller circle and AB is tangent to this circle at P,

? OP ? AB.
We know that, the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.

So, OP ? AB and AP = BP.

In ? APO, we have

OA2 = AP2 + OP2 => 52 = AP2 + 32

AP2 = 52 ? 32 = 25 ? 9 = 16
AP = 4

Now, AB = 2AP ???? [? AP = BP ]

AB= 2 × 4 = 8

? The length of the chord of the larger circle which touches the smaller circle is 8 cm.

Q8 ) A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :


? Lengths of two tangents drawn from an external point of circle are equal,

AP = AS , BP = BQ , DR = DS , and CR = CQ

Adding these all, we get

(AP + BP) + (CR + RD) 
= (BQ + QC) + (DS + SA)

=> AB + CD = BC + DA

Hence proved

Q9 ) In 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°.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :


Since tangents drawn from an external point to a circle are equal,

? AP = AC.
Thus in ? PAO and ? AOC, we have

AP = AC
AO = AO [Common]
PO = OC [Radii of the same circle]
? PAO ? AOC [ By SSS-criterion of congruence]

? PAO = ? CAO 
? ? PAC = 2?CAO

Similarly, we can prove that

? QBO = ? CBO 
? ? CBQ = 2? CBO

Now, ? PAC + ? CBQ = 180°
[Sum of the interior angles on the same side of transversal is 180°]

?2? CAO + 2? CBO = 180°

?? CAO + ? CBO = 90°
?180° ? ? AOB = 90°
[? ? CAO + ? CBO + ? AOB = 180°]

? ? 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 center.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :


Let PA and PB be two tangents drawn from an external point P to a circle with centre O. We have to prove that- ? AOB + ? APB = 180°



In right ? OAP and OBP, we have

PA = PB
[Tangents drawn from an external point are equal]

OA = OB
[Radii of the same circle]

OP = OP [Common]

?? OAP ? ? OBP
[by SSS - criterion of congruence]

?? OPA = ? OPB

and, ? AOP = ? BOP 
? ? APB = 2? OPA

and, ? AOB = 2? AOP

But, ? AOP = 90° ? ? OPA
[ ?? OAP is right triangle]

?2? AOP = 180° ? 2? OPA

?? AOB = 180° ? ? APB

?? AOB + ? APB = 180°

Hence proved 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.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :


Let ABCD be a parallelogram such that its sides touch a circle with centre O.



We know that the tangents to a circle from an exterior point are equal in length.

AP = AS , BP = BQ , DR = DS , and CR = CQ

Adding these all, we get

(AP + BP) + (CR + RD) = (BQ + QC) + (DS + SA)

? AB + CD = BC + DA

? 2AB = 2BC [? ABCD is a parallelogram ]
?AB = CD and BC = AD]

?AB = BC

? AB = BC = CD = AD

? 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 figure). Find the sides AB and AC.



NCERT Solutions for Class 10 Maths Chapter 10 Circles


Answer :


Let us join AO, OC, and OB.



It is given that BD = 8cm, CD = 6 cm.

? Lengths of two tangents drawn from an external point of circle are equal.

? BF = BD = 8cm ,
 CE = CD = 6cm ,
 and let AF = AE = x cm.

Then, the sides of the triangle are 14 cm, (x+6) cm and (x+8) cm.

?2s = 14 + (x+6) + (x+8) = 28 + 2x

? s = 14 + x

s ? a = 14 + x ? 14 = x,

s ? b = 14 + x ? x ? 6 = 8

and, s ? c = 14 + x ? x ? 8 = 6

? Area of ? ABC 
= s(s ? a)(s ? b)(s ? c)
= (14 + x)(x)(8)(6)

= 48(x2 + 14x)

Also, Area of ? ABC 
= Area(?OBC) + Area(?OCA) + Area(?OAB)

= 12 × BC × OD + 12 × CA × OE + 12 × AB × OF

= 12 × 14 × 4 + 12 × (x+6) × 4 + 12 × (x+8) × 4

= 2(14 + x + 6 + x + 8) = 2(28 + 2x)

?(48(x2 + 14x) = 4(14 + x)

Squaring, we get

48(x2 + 14x) = 16(14 + x)2

? 3(x2 + 14x) = 196 + 28x + x2

?2x2 + 14x ? 196 = 0

? x2 + 7x ? 98 = 0

? (x ? 7)(x + 14) = 0

? x = 7 or x = ?14

But x cannot be negative,

?x = 7

Thus, AB = x + 8 = 7 + 8 = 15 cm and AC = x + 6 = 6 + 7 = 13 cm.

Hence, the sides AB and AC are 15 cm and 13 cm respectively.

Q13 ) Prove that 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


Answer :


Let a circle with centre O touch the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively.

We have to prove that-

? AOB + ? COD= 180°

and, ? AOD + ? BOC = 180°
Join OP, OQ, OR and OS.

? The two tangents drawn from an external point to a circle subtend equal angles at the centre.



? ? 1 = ? 2 , ? 3 = ? 4 , ? 5 = ? 6 and ? 7 = ? 8

Now, ? 1 + ? 2 + ? 3 + ? 4 + ? 5 + ? 6 + ? 7 + ? 8 = 360° [? Sum of all the ? s around a point is 360°]

? 2(? 2 + ? 3 + ? 6 + ? 7) = 360°

and, 2(? 1 + ? 8 + ? 4 + ? 5) = 360°

? (? 2 + ? 3 + ? 6 + ? 7) = 180°

and (? 1 + ? 8) + (? 4 + ? 5) = 180°

Now since ?2+?3=?AOB ,
 ?6+?7=?COD,
 ?1+?8=?AOD  and  ?4+?5=?BOC, we can say that

? AOB + ? COD = 180°
and, ? AOD + ? BOC = 180°

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



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