## RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2

**Other Exercises**

- RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1
- RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2
- RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS
- RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

**Question 1.**

If PT is a tangent at T to a circle whose centre is O and OP = 17 cm, OT = 8 cm. Find the length of the tangent segment PT.

**Solution:**

PT is the tangent to the circle with centre O, at T

Radius OT = 8 cm, OP = 17 cm

PT is the tangent segment

Now in right ∆OPT,

OP² = OT² + PT² (Pythagoras Theorem)

=> (17)² = (8)² + PT²

=> 289 = 64 + PT²

=> PT² = 289 – 64 = 225 = (15)²

PT = 15 cm

**Question 2.**

Find the length of a tangent drawn to a circle with radius 5 cm, from a point 13 cm from the centre of the circle.

**Solution:**

From a point P outside the circle with centre O, PT is the tangent to the circle and radius

OT = 5 cm, OP = 15 cm

OT ⊥ PT

Now in right ∆OPT,

OP² = OT² + PT² (Pythagoras Theorem)

(13)² = (5)² + PT²

=> 169 = 25 + PT²

=> PT² = 169 – 25 = 144 = (12)²

PT = 12 cm

**Question 3.**

A point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle.

**Solution:**

From a point P outside the circle of centre 0 and radius OT, PT is the tangent to the circle

OP = 26 cm, PT = 10 cm

Now in right ∆OPT

Let r be the radius

OP² = OT² + PT² (Pythagoras Theorem)

=> (26)² = r² + (10)²

=> 676 = r² + 100

=> 676 – 100 = r²

=> r² = 576 = (24)²

r = 24

Hence radius of the circle = 24 cm

**Question 4.**

If from any point on the common chord of two intersecting circles, tangents be drawn to the circles, prove that they are equal.

**Solution:**

Given : QR is the common chord of two circles intersecting each other at Q and R

P is a point on RQ when produced From PT and RS are the tangents drawn to tire circles with centres O and C respectively

To prove : PT = PS

Proof: PT is the tangent and PQR is the secant to the circle with centre O

PT² = PQ x PR ….(i)

Similarly PS is the tangent and PQR is the secant to the circle with centre C

PS² = PQ x PR ….(ii)

From (i) and (ii)

PT² = PS²

PT = PS

Hence proved.

**Question 5.**

If the sides of a quadrilateral touch a circle, prove that the sum of a pair of opposite sides is equal to the sum of the other pair.

**Solution:**

Given : The sides of a quadrilateral ABCD touch the circle at P, Q, R and S respectively

To prove : AB + CD = AP + BC

Proof : AP and AS are the tangents to the circle from A

AP = AS ….(i)

Similarly BP = BQ ……(ii)

CR = CQ ….(iii)

and DR = DS ….(iv)

Adding, we get

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

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

=> AB + CD = AD + BC

Hence proved.

**Question 6.**

Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle. **[NCERT Exemplar]**

**Solution:**

Let C_{1} and C_{2} be the two circles having same centre O. AC is a chord which touches the C_{1} at point D

Join OD.

Also, OD ⊥ AC

AD = DC = 4 cm

[perpendicular line OD bisects the chord]

In right angled ∆AOD,

OA² = AD² + DO²

[by Pythagoras theorem, i.e.,

(hypotenuse)² = (base)² + (perpendicular)²]

=> DO² = 5² – 4² = 25 – 16 = 9

=> DO = 3 cm

Radius of the inner circle OD = 3 cm

**Question 7.**

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ. **[NCERT Exemplar]**

**Solution:**

Given : Chord PQ is parallel tangent at R.

To prove : R bisects the arc PRQ.

Proof: ∠1 = ∠2 [alternate interior angles]

∠1 = ∠3

[angle between tangent and chord is equal to angle made by chord in alternate segment]

∠2 = ∠3

=> PR = QR

[sides opposite to equal angles are equal]

=> PR = QR

So, R bisects PQ.

**Question 8.**

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A. **[NCERT Exemplar]**

**Solution:**

Given, AB is a diameter of the circle.

A tangent is drawn from point A.

Draw a chord CD parallel to the tangent MAN.

So, CD is a chord of the circle and OA is a radius of the circle.

∠MAO = 90°

[Tangent at any point of a circle is perpendicular to the radius through the point of contact]

∠CEO = ∠MAO [corresponding angles]

∠CEO = 90°

Thus, OE bisects CD

[perpendicular from centre of circle to chord bisects the chord]

Similarly, the diameter AB bisects all. Chord which are parallel to the tangent at the point A.

**Question 9.**

If AB, AC, PQ are the tangents in the figure, and AB = 5 cm, find the perimeter of ∆APQ.

**Solution:**

Given : AB, AC and PQ are the tangents to the circle as shown in the figure above and AB = 5 cm

To find : The perimeter of ∆APQ

Proof: PB and PX are the tangents to the circle

PB = PX

Similarly QC and QX are the tangents from

QC = QX

and AB and AC are the tangents from A

AB = AC

Now perimeter of ∆APQ

= AP + PQ + AQ

= AP + PX + QX + AQ

= AP + PB + QC + AQ { PB = PX and QC = QX}

= AB + AC

= AB + AB (AB=AC)

= 2 AB = 2 x 5 = 10 cm

**Question 10.**

Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

**Solution:**

Given : PQ and RS are parallel tangents of a circle

RMP is the intercept of the tangent between PQ and RS

RO and PQ are joined

**Question 11.**

In the figure, PQ is tangent at a point R of the circle with centre O. If ∠TRQ = 30°, find m ∠PRS

**Solution:**

In the figure,

PRQ is tangent to the circle with centre O at R

RT and RS are joined such that ∠TRQ = 30°

Let ∠PRS = x°

Now ∠SRX = 90° (angle in a semicircle)

But ∠TRQ + ∠SRT + ∠PRS = 180° (Angles of a line)

=> 30° + 90° + x° = 180°

=> 120° + x° = 180°

=> x° = 180° – 120° = 60°

∠PRS = 60°

**Question 12.**

If PA and PB are tangents from an outside point P, such that PA = 10 cm and ∠APB = 60°. Find the length of chord AB.

**Solution:**

PA and PB are the tangents from a point PQ outside the circle with centre O

PA = 10 cm and ∠APB = 60°

Tangents drawn from a point outside the circle are equal

PA = PB = 10 cm ∠PAB = ∠PBA

(Angles opposite to equal sides)

But in ∆APB,

∠APB + ∠PAB + ∠PBA = 180° (Angles of a triangle)

=> 60° + ∠PAB + ∠PAB = 180°

=> 2 ∠PAB = 180° – 60° = 120°

∠PAB = 60°

∠PBA = ∠PAB = 60°

PA = PB = AB = 10 cm

Hence length of chord AB = 10 cm

**Question 13.**

In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC. **[NCERT Exemplar]**

**Solution:**

Let O be the centre of the given circle. Suppose, the tangent at P meets BC at Q.

Join BP.

To prove : BQ = QC

[angles in alternate segment]

Proof : ∠ABC = 90°

[tangent at any point of circle is perpendicular to radius through the point of contact]

In ∆ABC, ∠1 + ∠5 = 90°

[angle sum property, ∠ABC = 90°]

∠3 = ∠1

[angle between tangent and the chord equals angle made by the chord in alternate segment]

∠3 + ∠5 = 90° ……..(i)

Also, ∠APB = 90° [angle in semi-circle]

∠3 + ∠4 = 90° …….(ii)

[∠APB + ∠BPC = 180°, linear pair]

From Eqs. (i) and (ii), we get

∠3 + ∠5 = ∠3 + ∠4

∠5 = ∠4

=> PQ = QC

[sides opposite to equal angles are equal]

Also, QP = QB

[tangents drawn from an internal point to a circle are equal]

=> QB = QC

Hence proved.

**Question 14.**

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ∆PCD.

**Solution:**

PA and PB are the tangents drawn from a point P out side the circle with centre O

CD is another tangents to the circle at point E which intersects PA and PB at C and D respectively

PA = 14 cm

PA and PB are the tangents to the circle from P

PA = PB = 14 cm

Now CA and CE are the tangents from C

CA = CE ….(i)

Similarly DB and DE are the tangents from D

DB = DE ….(ii)

Now perimeter of ∆PCD

= PC + PD + CD

= PC + PD + CE + DE

= PC + CE + PD + DE

= PC + CA + PD = DB {From (i) and (ii)}

= PA + PB

= 14 + 14

= 28 cm

**Question 15.**

In the figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle. **[CBSE 2002]**

**Solution:**

In right ∆ABC, ∠B = 90°

BC = 6 cm, AB = 8 cm

Let r be the radius of incircle whose centre is O and touches the sides A B, BC and CA at P, Q and R respectively

AP and AR are the tangents to the circle AP = AR

Similarly CR = CQ and BQ = BP

OP and OQ are radii of the circle

OP ⊥ AB and OQ ⊥ BC and ∠B = 90° (given)

BPOQ is a square

BP = BQ = r

AR = AP = AB – BD = 8 – r

and CR = CQ = BC – BQ = 6 – r

But AC² = AB² + BC² (Pythagoras Theorem)

= (8)² + (6)² = 64 + 36 = 100 = (10)²

AC = 10 cm

=> AR + CR = 10

=> 8 – r + 6 – r = 10

=> 14 – 2r = 10

=> 2r = 14 – 10 = 4

=> r = 2

Radius of the incircle = 2 cm

**Question 16.**

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. **[NCERT Exemplar]**

**Solution:**

Let mid-point of an arc AMB be M and TMT’ be the tangent to the circle.

Join AB, AM and MB.

Since, arc AM = arc MB

=> Chord AM = Chord MB

In ∆AMB, AM = MB

=> ∠MAB = ∠MBA ……(i)

[equal sides corresponding to the equal angle]

Since, TMT’ is a tangent line.

∠AMT = ∠MBA

[angle in alternate segment are equal]

∠AMT = ∠MAB [from Eq. (i)]

But ∠AMT and ∠MAB are alternate angles, which is possible only when AB || TMT’

Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Hence proved

**Question 17.**

From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle show that ∆APB is equilateral.

**Solution:**

Given : From a point P outside the circle with centre O, PA and PB are the tangerts to the circle such that OP is diameter.

AB is joined.

To prove: APB is an equilateral triangle

Const : Join OP, AQ, OA

Proof : OP = 2r

=> OQ + QP = 2r

=> OQ = QP = r (OQ = r)

Now in right ∆OAP,

OP is its hypotenuse and Q is its mid point

OA = AQ = OQ

(mid-point of hypotenuse of a right triangle is equidistances from its vertices)

∆OAQ is equilateral triangle ∠AOQ = 60°

Now in right ∆OAP,

∠APO = 90° – 60° = 30°

=> ∠APB = 2 ∠APO = 2 x 30° = 60°

But PA = PB (Tangents from P to the circle)

=> ∠PAB = ∠PBA = 60°

Hence ∆APB is an equilateral triangle.

**Question 18.**

Two tangents segments PA and PB are drawn to a circle with centre O such that ∠APB = 120°. Prove that OP = 2 AP. **[CBSE 2014]**

**Solution:**

Given : From a point P. Out side the circle with centre O, PA and PB are tangerts drawn and ∠APB = 120°

OP is joined To prove : OP = 2 AP

Const: Take mid point M of OP and join AM, join also OA and OB.

Proof : In right ∆OAP,

∠OPA = \(\frac { 1 }{ 2 }\) ∠APB = \(\frac { 1 }{ 2 }\) x 120° = 60°

∠AOP = 90° – 60° = 30°

M is mid point of hypotenuse OP of ∆OAP

MO = MA = MP

∠OAM = ∠AOM = 30° and ∠PAM = 90° – 30° = 60°

∆AMP is an equilateral triangle

MA = MP = AP

But M is mid point of OP

OP = 2 MP = 2 AP

Hence proved.

**Question 19.**

If ∆ABC is isosceles with AB = AC and C (0, r) is the incircle of the ∆ABC touching BC at L. Prove that L bisects BC.

**Solution:**

Given: In ∆ABC, AB = AC and a circle with centre O and radius r touches the side BC of ∆ABC at L.

To prove : L is mid point of BC.

Proof : AM and AN are the tangents to the circle from A

AM = AN

But AB = AC (given)

AB – AN = AC – AM

BN = CM

Now BL and BN are the tangents from B

BL = BN

Similarly CL and CM are tangents

CL = CM

But BM = CM (proved)

BL = CL

L is mid point of BC.

**Question 20.**

AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD. **[NCERT Exemplar]**

**Solution:**

To prove, BC = BD

Join BC and OC.

Given, ∠BAC = 30°

=> ∠BCD = 30°

[angle between tangent and chord is equal to angle made by chord in the alternate segment]

∠ACD = ∠ACO + ∠OCD

∠ACD = 30° + 90° = 120°

[OC ⊥ CD and OA = OC = radius => ∠OAC = ∠OCA = 30°]

In ∆ACD,

∠CAD + ∠ACD + ∠ADC = 180°

[since, sum of all interior angles of a triangle is 180°]

=> 30° + 120° + ∠ADC = 180°

=> ∠ADC = 180° – (30° + 120°) = 30°

Now, in ∆BCD,

∠BCD = ∠BDC = 30°

=> BC = BD

[since, sides opposite to equal angles are equal]

**Question 21.**

In the figure, a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm, and CD = 4 cm. Find AD. **[CBSE 2002]**

**Solution:**

A circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at P, Q, R and S respectively.

AB = 6 cm, BC = 7 cm, CD = 4cm

Let AD = x

AP and AS are the tangents to the circle

AP = AS

Similarly,

BP = BQ

CQ = CR

and OR = DS

AB + CD = AD + BC

=> 6 + 4 = 7 + x

=> 10 = 7 + x

=>x = 10 – 7 = 3

AD = 3 cm

**Question 22.**

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

**Solution:**

Given : TS is a tangent to the circle with centre O at P, OP is joined

To prove : OP is perpendicular to TS which passes through the centre of the circle

Construction : Draw a line OR which intersect the circle at Q and meets the tangent TS at R

Proof: OP = OQ

(radii of the same circle) and OQ < OR => OP < OR

Similarly we can prove that OP is less than all lines which can be drawn from O to TS

OP is the shortest

OP is perpendicular to TS

Perpendicular through P, will pass through the centre of the circle

Hence proved.

**Question 23.**

Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. Prove that TQ = TR.

**Solution:**

Given : Two circles with centres O and C touch each other externally at P. PT is its common tangent

From a point T on PT, TR and TQ are the tangents drawn to the circles

To prove : TQ = TR

Proof : From T, TR and TP are two tangents to the circle with centre O

TR = TP ….(i)

Similarly, from T,

TQ and TP are two tangents to the circle with centre C

TQ = TP ….(ii)

From (i) and (ii)

TQ = TR

Hence proved.

**Question 24.**

A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC. **[NCERT Exemplar]**

**Solution:**

Given : Two tangents are drawn from an external point A to the circle with centre O, Tangent BC is drawn at a point R, radius of circle equals to 5 cm.

To find : Perimeter of ∆ABC.

Proof : ∠OPA = 90°

[Tangent at any point of a circle is perpendicular to the radius through the point of contact]

OA² = OP² + PA² [by Pythagoras Theorem]

(13)² = 5² + PA²

=> PA² = 144 = 12²

=> PA = 12 cm

Now, perimeter of ∆ABC = AB + BC + CA = (AB + BR) + (RC + CA)

= AB + BP + CQ + CA [BR = BP, RC = CQ tangents from internal point to a circle are equal]

= AP + AQ = 2AP = 2 x (12) = 24 cm

[AP = AQ tangent from internal point to a circle are equal]

Hence, the perimeter of ∆ABC = 24 cm.

**Question 25.**

In the figure, a circle is inscribed in a quadrilateral ABCD in which ∠B = 90°. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius r of the circle.

**Solution:**

In the figure, O is the centre of the circle inscribed in a quadrilateral ABCD and ∠B = 90°

AD = 23 cm, AB = 29 cm, DS = 5 cm

OP = OQ (radii of the same circle)

AB and BC are tangents to the circle and OP and OQ are radii

OP ⊥ BC and OQ ⊥ AB

∠OPB = ∠OQB = 90°

PBQO is a square

DS and DR are tangents to the circle

DR = DS = 5 cm

AR = AD – DR = 23 – 5 = 18 cm

AR and AQ are the tangents to the circle

AQ = AR = 18 cm But AB = 29 cm

BQ = AB – AQ = 29 – 18 = 11 cm

Side of square PBQO is 11 cm

OP = 11 cm

Hence radius of the circle = 11 cm

**Question 26.**

In the figure, there are two concentric circles with centre O of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If AP =12 cm, find the length of BP. **[CBSE 2010]**

**Solution:**

Two concentric circles with centre O with radii 5 cm and 3 cm respectively from a

point P, PA and BP are tangents drawn to there circles

AP = 12 cm

To find BP

In right ∆OAP,

OP² = OA² + AP² (Pythagoras Theorem)

= (5)² + (12)² = 25 + 144

= 169 = (13)²

OP = 13 cm

Now in right ∆OBP,

OP² = OB² + BP²

=> (13)² = (3)² + BP²

=> 169 = 9 + BP²

=> BP² = 169 – 9 = 160 = 16 x 10

BP = √(16 x 10) = 4√10 cm

**Question 27.**

In the figure, AB is a chord of length 16 cm of a circle of radius 10 cm. The tangents at A and B intersect at a point P. Find the length of PA. **[CBSE 2010]**

**Solution:**

In the figure, AB is the chord of the circle with centre O and radius 10 cm.

Two tangents from P are drawn to the circle touching it at A and B respectively

AB is joined with intersects OP at L

**Question 28.**

In the figure, PA and PB are tangents from an external point P to a circle with centre O. LN touches the circle at M. Prove that PL + LM = PN + MN. **[CBSE 2010]**

**Solution:**

Given : In the figure, PA and PB are the tangents to the circle with centre O from a point P outside it LN touches it at M

To prove : PL + LM = PN + MN

Prove : PA and PB are tangents to the circle from P

PA = PB

Similarly from L, LA and LM are tangents

LA = LM

Similarly NB = NM

Now PA = PB => PL + LA = PN + NB

PL + LM = PN + NM

Hence proved.

**Question 29.**

In the figure, BDC is a tangent to the given circle at point D such that BD = 30 to the circle and meet when produced at A making BAC a right angle triangle. Calculate (i) AF (ii) radius of the circle.

**Solution:**

In the figure, BDC is a tangent to the given circle with centre O and D is a point such that

BD = 30 cm and CD = 7 cm

BE and CF are other two tangents drawn from B and C respectively which meet at A on producing this and ∆BAC is a right angle so formed

To find : (i) AF and (ii) radius of the circle

Join OE and OF

OE = OF radii of the circle

OE ⊥ AB and OF ⊥ AC

OEAF is a square

BD and BE are the tangents from B

BE = BD = 30 cm and similarly

CF = CD = 7 cm

Let r be the radius of the circle

OF = AF = AE = r

AB = 30 + r and AC = 7 + r and BC = 30 + 7 = 37 cm

Now in right ∆ABC

**Question 30.**

If d_{1}, d_{2} (d_{2} > d_{1}) be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to the other circle, prove that \({ d }_{ 2 }^{ 2 }={ c }^{ 2 }+{ d }_{ 1 }^{ 2 }\). **[NCERT Exemplar]**

**Solution:**

Let AB be a chord of a circle which touches the other circle at C. Then ∆OCB is right triangle.

**Question 31.**

In the given figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS.** [CBSE 2015, NCERT Exemplar]**

**Solution:**

In the given figure,

PQ and PR are tangents to the circle with centre O drawn from P

∠RPQ = 30°

Chord RS || PQ is drawn

To find ∠RQS

PQ = PR (tangents to the circle)

∠PRQ = ∠PQR But ∠RPQ = 30°

**Question 32.**

From an external point P, tangents PA = PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB. **[CBSE 2016]**

**Solution:**

PA = PB [tangents drawn from external point are equal]

∠PBA = ∠PAB = 50° [angles equal to opposite sides]

∠APB = 180° – 50° – 50° = 80° [angle-sum property of a A]

In cyclic quad. OAPB

∠AOB + ∠APB = 180° [sum of opposite angles of a cyclic quadrilateral is 180°]

∠AOB + 80° = 180°

∠AOB = 180°- 80° = 100°

**Question 33.**

In the figure, two tangents AB and AC are drawn to a circle with centre O such that ∠BAC = 120°. Prove that OA = 2AB.

**Solution:**

Given : In the figure, O is the centre of the circle.

AB and AC are the tangents to the circle from A such that

∠BAC = 120° .

To prove : OA = 2AB

Proof : In ∆OAB and ∆OAC

∠OBA = ∠OCA – 90° (OB and OC are radii)

OA = OA (common)

OB = OC (radii of the circle)

∆OAB ~ ∆OAC

∠OAB = ∠OAC = 60°

Now in right ∆OAB,

**Question 34.**

The lengths of three concecutive sides of a quadrilateral circumscribing a circle are 4 cm, 5 cm, and 7 cm respectively. Determine the length of the fourth side.

**Solution:**

In quadrilateral ABCD which is circumerscribing it

BC = 4 cm, CD = 5 cm and DA = 7 cm

We know that if a quad, is circumscribed in a circle, then

AB + CD = AD + BC

=> AB + 5 = 4 + 7

=> AB + 5 = 11

AB = 11 – 5 = 6

AB = 6 cm

**Question 35.**

The common tangents AB and CD to two circles with centres O and O’ intersect at E between their centres. Prove that the points O, E and O’ are collinear. **[NCERT Exemplar]**

**Solution:**

Joint AO, OC and O’D, O’B

Now, in ∆EO’D and ∆EO’B

O’D = O’B [radius]

O’E = O’E [common side]

ED = EB

**Question 36.**

In the figure, common tangents PQ and RS to two circles intersect at A. Prove that PQ = RS.

**Solution:**

Given : Two common tangents PQ and RS intersect each other at A.

To prove : PQ = RS

Proof: From A, AQ and AR are two tangents are drawn to the circle with centre O.

AP = AR ….(i)

Similarly AQ and AS are the tangents to the circle with centre C

AQ = AS ….(ii)

Adding (i) and (ii)

AP + AQ = AR + AS

=> PQ = RS

Hence proved.

**Question 37.**

Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle. **[CBSE 2014]**

**Solution:**

Let R be the radius of outer circle and r be the radius if small circle of two concentric circle

AB is the chord of the outer circle and touches the smaller circle at P

Join OP, OA

**Question 38.**

AB and CD are common tangents to two circles of equal radii. Prove that AB = CD. **[NCERT Exemplar]**

**Solution:**

Given : AB and CD are tangents to two circles of equal radii.

To prove :

Construction : Join OA, OC, O’B and O’D

Proof: Now, ∠OAB = 90°

[tangent at any point of a circle is perpendicular to radius through the point of contact]

Thus, AC is a straight line.

Also, ∠OAB + ∠OCD = 180°

AB || CD

Similarly, BD is a straight line and ∠O’BA = ∠O’DC = 90°

Also, AC = BD

[radii of two circles are equal] In quadrilateral ABCD,

∠A = ∠B = ∠C = ∠D = 90°

andAC = BD

ABCD is a rectangle

Hence, AB = CD

[opposite sides of rectangle are equal]

**Question 39.**

A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆PQR is 336 cm², find the sides PQ and PR. **[CBSE 2014]**

**Solution:**

∆PQR is circumscribed by a circle with centre O and radius 8 cm

T is point of contact which divides the line segment OT into two parts such that

QT = 14 cm and TR = 16 cm

Area of ∆PQR = 336 cm²

Let PS = x cm

QT and QS are tangents to the circle from Q

QS = QT = 14 cm

Similarly RU and RT are tangents to the circle

RT = RU = 16 cm

Similarly PS and PU are tangents from P

PS = PU = x

Now PQ = x + 14 and PR = x + 16 and QR = 14 + 16 = 30 cm

Now area of ∆PQR = Area of ∆POQ + area of ∆QOR + area of ∆POR

**Question 40.**

In the figure, the tangent at a point C of a circle and a diameter AB when extended itersect at P. If ∠PCA = 110°, find ∠CBA. **[NCERT Exemplar]**

**Solution:**

Here, AB is a diameter of the circle from point C and a tangent is drawn which meets at a point P.

Join OC. Here, OC is radius.

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

OC ⊥ PC

Now, ∠PCA = 110° [given]

=> ∠PCO + ∠OCA = 110°

=> 90° + ∠OCA = 110°

=> ∠OCA = 20°

OC = OA = Radius of circle

∠OCA = ∠OAC = 20°

[since, two sides are equal, then their opposite angles are equal]

Since, PC is a tangent, so

∠BCP = ∠CAB = 20°

[angles in a alternate segment are equal]

In ∆PBC, ∠P + ∠C + ∠A= 180°

∠P = 180° – (∠C + ∠A)

∠P = 180° – (110° + 20°)

∠P = 180° – 130° = 50°

In ∆PBC,

∠BPC + ∠PCB + ∠PBC = 180°

[sum of all interior angles of any triangle is 180°]

=> 50° + 20° + ∠PBC = 180°

=> ∠PBC = 180° – 70°

∠PBC = 110°

Since, ∆PB is a straight line.

∠PBC + ∠CBA = 180°

∠CBA = 180° – 110° = 70°

**Question 41.**

AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in the figure. Prove that ∠BAT = ∠ACB. **[NCERT Exemplar]**

**Solution:**

Since, AC is a diameter line, so angle in semicircle makes an angle 90°.

∠ABC = 90° [by property]

In ∆ABC,

∠CAB + ∠ABC + ∠ACB = 180°

[ sum of all interior angles of any triangle is 180°]

=> ∠CAB + ∠ACB = 180° – 90° = 90° ……….(i)

Since, diameter of a circle is perpendicular to the tangent.

i.e. CA ⊥ AT

∠CAT = 90°

=> ∠CAB + ∠BAT = 90° …….(ii)

From Eqs. (i) and (ii),

∠CAB + ∠ACB = ∠CAB + ∠BAT

=> ∠ACB = ∠BAT

Hence proved.

**Question 42.**

In the given figure, a ∆ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of ∆ABC is 84 cm². **[CBSE 2015]**

**Solution:**

In the given figure,

In ∆ABC is circle is inscribed touching it at D, E and F respectively.

Radius of the circle (r) = 4cm

OD ⊥ BC, then

OD = 4 cm, BD = 8 cm, DC = 6 cm

Join OE and OF

**Question 43.**

In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ. **[CBSE 2015]**

**Solution:**

In the given figure,

AB is the diameter, AT is the tangent

and ∠AOQ = 58°

To find ∠ATQ

Arc AQ subtends ∠AOQ at the centre and ∠ABQ at the remaining part of the circle

∠ABQ = \(\frac { 1 }{ 2 }\) ∠AOQ = \(\frac { 1 }{ 2 }\) x 58° = 29°

Now in ∆ABT,

∠BAT = 90° ( OA ⊥ AT)

∠ABT + ∠ATB = 90°

=> ∠ABT + ∠ATQ = 90°

=> 29° + ∠ATQ = 90°

=> ∠ATQ = 90°- 29° = 61°

**Question 44.**

In the figure, OQ : PQ = 3:4 and perimeter of ∆POQ = 60 cm. Determine PQ, QR and OP.

**Solution:**

In the figure, OQ : PQ = 3:4

Perimeter of ∆POQ = 60 cm

To find PQ, QR and OP

OQ : PQ = 3 : 4

Let OQ = 3x and PQ = 4x

Now in right ∆OPQ,

OP² = OQ² + PQ² = (3x)² + (4x)² = 9x² + 16x² = 25x² = (5x)²

OP = 5x

But OQ + QP + OP = 60 cm

3x + 4x + 5x = 60

=> 12x = 60

x = 5

PQ = 4x = 4 x 5 = 20 cm

QR = 2 OQ = 2 x 3x = 6 x 5 = 30 cm

OP = 5x = 5 x 5 = 25 cm

**Question 45.**

Equal circles with centre O and O’ touch each other at X. OO’ produced to meet a circle with centre O’, at A. AC is a tangent to the circle whose centre is O. O’ D is perpendicular to AC. Find the value of \(\frac { DO’ }{ CO }\)

**Solution:**

Two equal circles with centre O and O’ touch each other externally at X

OO’ produced to meet at A

AC is the tangent of circle with centre O,

O’D ⊥ AC is drawn OC is joined

AC is tangent and OC is the radius

OC ⊥ AC

O’D ⊥ AC

OC || O’D

Now O’A = \(\frac { 1 }{ 2 }\) A x or \(\frac { 1 }{ 2 }\) AO

Now in O’AD and AOAC

∠A = ∠A (common)

∠AO’D = ∠AOC (corresponding angles)

**Question 46.**

In the figure, BC is a tangent to the circle with centre O. OE bisects AP. Prove that ∆AEO ~ ∆ABC.

**Solution:**

Given : In the figure, BC is a tangent to the circle with centre O at B.

AB is diameter AC is joined which intersects the circle at P

OE bisects AP

To prove : ∆AEO ~ ∆ABC

Proof: In ∆OAE and ∆OPE

OE = OE (common)

OA = OP ‘ (radii of the same circle)

EA = EP (given)

∆OAE = ∆OPE (SSS axiom)

∠OEA = ∠OEP

But ∠OEA + ∠OEP = 180°

∠OEA = 90°

Now in ∆AEO and ∆ABC

∠OEA = ∠ABC (each 90°)

∠A = ∠A (common)

∆AEO ~ ∆ABC (AA axiom)

Hence proved.

**Question 47.**

In the figure, PO ⊥ QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.

**Solution:**

Given : In the figure, O is the centre of the circle

PO ⊥ QO

They tangents at P and Q intersect each other at T

To prove : PQ and OT are right bisector of each other

Proof : PT and QT are tangents to the circle

PT = QT

OP and OQ are radii of the circle and ∠POQ = 90° ( PO ⊥ QO)

OQTP is a square Where PQ and OT are diagonals

Diagonals of a square bisect each other at right angles

PQ and OT bisect each other at right angles

Hence PQ and QT are right bisectors of each other.

**Question 48.**

In the figure, O is the centre of the circle and BCD is tangent to it at C. Prove that ∠BAC + ∠ACD = 90°.

**Solution:**

Given : In the figure, O is the centre of the circle BCD is a tangent, CP is a chord

prove : ∠BAC + ∠ACD = 90°

Proof: ∠ACD = ∠CPA (Angles in the alternate segment)

But in ∆ACP,

∠ACP = 90° (Angle in a semicircle)

∠PAC + ∠CPA = 90°

=> ∠BAC + ∠ACD = 90°

(∠ACD = ∠CPA proved)

Hence proved.

**Question 49.**

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines. **[NCERT Exemplar]**

**Solution:**

Given : Two tangents PQ and PR are drawn from an external point P to a circle with centre O.

To prove : Centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Construction : Join OR, and OQ.

In ∆POR and ∆POQ

∠PRO = ∠PQO = 90°

[tangent at any point of a circle is perpendicular to the radius through the point of contact]

OR = OQ [radii of same circle]

Since, OP is common.

∆PRO = ∆PQO [RHS]

Hence, ∠RPO = ∠QPO [by CPCT]

Thus, O lies on angle bisector of PR and PQ.

Hence proved.

**Question 50.**

In the figure, there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the lengths of PS. **[CBSE 2017]**

**Solution:**

Construction : Join OS and OP.

Consider ∆POS

We have,

PO = OS

∆POS is an isosceles triangle.

We know that in an isosceles triangle, if a line drawn perpendicular to the base of the triangle from the common vertex of the equal sides, then that line will bisect the base (unequal side).

And PQ = PR = 5 cm

[PRT and PQS are tangents to the inner circle to the inner circle from a point P lying on the outer circle]

We have, PQ = QS

It is given that, PQ = 5 cm

QS = 5 cm

From the figure, we have

PS = PQ + QS

=> PS = 5 + 5

=> PS = 10 cm

**Question 51.**

In the figure, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOP is a diameter. If ∠POR = 130° and S is a point on the circle, find ∠1 + ∠2. **[CBSE 2017]**

**Solution:**

Construction : Join RT.

Given, ∠POR = 130°

∠POQ = 180°- (∠POR) = 180° – 130° = 50°

Since, PQ is a tangent

∠PQO = 90°

Now, In ∆POQ,

∠POQ + ∠PQO + ∠QPO = 180°

=> 50° + 90° + ∠1 = 180°

=> ∠1 = 180° – 140°

=> ∠1 = 40°

Now, In ∆RST

∠RST = \(\frac { 1 }{ 2 }\) ∠ROT

[Angle which is subtended by on arc at the centre of a circle is double the size of the angle subtended at any point on the circumference]

=> ∠2 = \(\frac { 1 }{ 2 }\) x 130° = 65°

Now ∠1 + ∠2 = 40° + 65° = 105°

**Question 52.**

In the figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC = PD. **[CBSE 2017]**

**Solution:**

PA = PB = 12 cm …(i)

QC = AC = 3cm …(ii)

QD = BD = 3 cm …(iii)

[Tangents drawn from an external point are equal]

To find : PC + PD

= (PA – AC) + (PB – BD)

= (12 – 3) + (12 – 3) [From (i), (ii), and (iii)]

= 9 + 9 = 18 cm

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2 are helpful to complete your math homework.

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