# ICSE Solutions for Chapter 18 Tangents and Intersecting Chords Class 10 Mathematics

**Question 1: Prove that of any two chord of a circle, the greater chord is nearer to the centre.**

**Solution 1:**

__Given:__A circle with centre and radius r. OM ⊥ AB and ON ⊥ CD. Also AB > CD

__To prove:__OM < ON

__Proof:__Join OA and OC.

In Rt. Î”AOM,

AO

^{2}= AM

^{2}+ OM

^{2}

⇒ r

^{2}= (1/2AB)

^{2}+ OM

^{2}

⇒ r

^{2}= 1/4AB

^{2}+ OM

^{2}

**...(i)**

Again in Rt. Î”ONC,

OC

^{2}= NC

^{2}+ ON

^{2}

⇒ r

^{2}= (1/2 OD)

^{2}+ ON

^{2}

⇒ r

^{2}= (1/4OD)

^{2}+ ON

^{2}

**...(ii)**

From (i) and (ii)

1/4AB

^{2}+ OM

^{2}= 1/4CD

^{2}+ ON

^{2}

But, AB > CD

**(given)**

∴ ON > OM

⇒ OM < ON

Hence, AB is nearer to the centre than CD.

**Question 2: In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm.**

**Find**

**(i) AB.**

**(ii) the length of tangent PT.**

**Solution 2:**

(i) PA = AB + BP = (AB + 4) cm

PC = PD + CD = 5 + 7.8 = 12.8 cm

Since PA × PB = PC x PD

⇒ (AB + 4) × 4 = 12.8 ×5

⇒ (AB + 4) = (12.8 × 5)/4

⇒ AB + 4 = 16

⇒ AB = 12 cm

(ii) Since PT

⇒ PT

⇒ PT

⇒ PT = 8 cm

∠QAB = ∠ADB = 30°

^{2}= PC × PD⇒ PT

^{2}= 12.8× 5⇒ PT

^{2}= 64⇒ PT = 8 cm

**Question 3: In the following figure, PQ is the tangent to the circle at A. DB is a diameter and O is the centre of the circle. If****∠ADB = 30° and ∠CBD = 60°: calculate:****(i) ∠QAB****(ii) ∠PAD****(iii) ∠CDB**

**Solution 3:**(i) PAQ Is a tangent and AB is the chord.∠QAB = ∠ADB = 30°

**(angles in the alternate segment)**

(ii) OA - OD

∴ ∠OAD = ∠ODA = 30°

But, OA ⊥ PQ

∴ ∠PAD - ∠OAP - ∠OAD = 90° - 30° = 60°

**(radli of the same circle)**∴ ∠OAD = ∠ODA = 30°

But, OA ⊥ PQ

∴ ∠PAD - ∠OAP - ∠OAD = 90° - 30° = 60°

(iii) BD is the diameter.

∴ ∠BCD = 90°

Now in Î”BCD,

∠CDB + ∠CBD + ∠BCD = 180°

⇒ ∠CDB + 60° + 90° = 180°

⇒ CDB = 180° - 150° = 30°

Therefore, QA – QP

Similarly, from Q, QB and QP are two tangents to the circle with centre O’.

Therefore, QB = QP

From (i) and (ii)

QA = QB

Therefore, tangents QA and QB are equal.

OS = 5 cm

OT = 3 cm

In Rt. Triangle OST

By Pythagoras Theorem,

ST

ST

ST

ST = 4 cm

Since OT is perpendicular to SP and OT bisects chord SP

So, SP = 8 cm

AB = 6 cm, AC = 8 cm and BC = 9 cm

Let radii of the circles having centers A, B and C be r

r

r

r

Adding

r

2(r

r

r

r

r

r

r

r

Hence, r

Since AP and AS are tangents to the circle from external point A

AP = AS .

Similarly, we can prove that:

BP = BQ

CR = CQ

DR = DS

Adding,

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

AB + CD = AD + BC

Hence, AB + CD = AD + BC

Since TPS is the tangent, PD is the chord.

∴ ∠PAB = ∠BPS

Similarly,

∠PCD = ∠DPS

Subtracting (i) from (ii)

∠PCD - ∠PAB = ∠DPS - ∠BPS

But in Î”PAC,

Ext. ∠PCD = ∠PAB + ∠CPA

∴ ∠PAB + ∠CPA - ∠PAB = ∠DPS - ∠BPS

⇒ ∠CPA = ∠DPB

Similarly,

∠ABD = ∠ACD

But, ∠AOB = ∠ACD

∴ ∠ADB = ∠ABD

TAS is a tangent and AB is a chord

∴ ∠BAS = ∠ADB

But, ∠ADB = ∠ABD

∴ ∠BAS = ∠ABD

But these are alternate angles

Therefore, TS∥BD.

Since, from A, AP and AR are the tangents to the circle

Therefore, AP = AR

Similarly, we can prove that

BP = BQ and CR = CQ

Adding,

AP + BP + CQ = AR + BQ + CR

(AP + BP) + CQ = (AR + CR) + BQ

AB + CQ = AC + BQ

But AB = AC

Therefore, CQ = BQ or BQ = CQ

∴ ∠BCD = 90°

**(angle in a semi-circle)**Now in Î”BCD,

∠CDB + ∠CBD + ∠BCD = 180°

⇒ ∠CDB + 60° + 90° = 180°

⇒ CDB = 180° - 150° = 30°

**Question 4: Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.**

**Solution 4:****From Q, QA and QP are two tangents to the circle with centre O**

Therefore, QA – QP

**...(i)**Similarly, from Q, QB and QP are two tangents to the circle with centre O’.

Therefore, QB = QP

**...(ii)**From (i) and (ii)

QA = QB

Therefore, tangents QA and QB are equal.

**Question 5: Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.**

**Solution 5:**OT = 3 cm

In Rt. Triangle OST

By Pythagoras Theorem,

ST

^{2}= OS^{2}– OT^{2}ST

^{2}= 25 - 9ST

^{2}= 16ST = 4 cm

Since OT is perpendicular to SP and OT bisects chord SP

So, SP = 8 cm

**Question 6: Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.**

**Solution 6:**Let radii of the circles having centers A, B and C be r

_{1}, r_{2}and r_{3}respectively.r

_{1}+ r_{3}= 8r

_{3}+ r_{2}= 9r

_{2}+ r_{1}= 6Adding

r

_{1}+ r_{3}+ r_{3}+ r_{2}+ r_{2}+ r_{1}= 8 + 9 + 62(r

_{1}+ r_{2}+ r_{3}) = 23r

_{1}+ r_{2}+ r_{3}= 11.5 cmr

_{1}+ 9 = 11.5 (Since r_{2}+ r_{3}= 9)r

_{1}= 2.5 cmr

_{2}+ 6 = 11.5 (Since r_{1}+ r_{3}= 6)r

_{2}= 5.5 cmr

_{3}+ 8 = 11.5 (Since r_{2}+ r_{1}= 8)r

_{3}= 3.5 cmHence, r

_{1}= 2.5 cm, r_{2}= 5.5 cm and r_{3}= 3.5 cm

**Question 7: If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.**

**Solution 7:****Let the circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.**

Since AP and AS are tangents to the circle from external point A

AP = AS .

**...(i)**Similarly, we can prove that:

BP = BQ

**...(ii)**CR = CQ

**...(iii)**DR = DS

**...(iv)**Adding,

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

AB + CD = AD + BC

Hence, AB + CD = AD + BC

**Question 8: Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠CPA = ∠DPB****Solution 8:****Draw a tangent TS at P to the circles given.**

Since TPS is the tangent, PD is the chord.

∴ ∠PAB = ∠BPS

**...(i) (angles in alternate segment)**Similarly,

∠PCD = ∠DPS

**...(ii)**Subtracting (i) from (ii)

∠PCD - ∠PAB = ∠DPS - ∠BPS

But in Î”PAC,

Ext. ∠PCD = ∠PAB + ∠CPA

∴ ∠PAB + ∠CPA - ∠PAB = ∠DPS - ∠BPS

⇒ ∠CPA = ∠DPB

**Question 9: In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.**

**Solution 9:****∠ADB = ∠ACB**

**...(i) (angles in same segment)**Similarly,

∠ABD = ∠ACD

**...(ii)**But, ∠AOB = ∠ACD

**(AC is bisector of ∠BCD)**∴ ∠ADB = ∠ABD

**(from i and ii)**TAS is a tangent and AB is a chord

∴ ∠BAS = ∠ADB

**(angles in alternate segment)**But, ∠ADB = ∠ABD

∴ ∠BAS = ∠ABD

But these are alternate angles

Therefore, TS∥BD.

**Question 10: In the figure, if AB = AC then prove that BQ = CQ.**

**Solution 10:**Since, from A, AP and AR are the tangents to the circle

Therefore, AP = AR

Similarly, we can prove that

BP = BQ and CR = CQ

Adding,

AP + BP + CQ = AR + BQ + CR

(AP + BP) + CQ = (AR + CR) + BQ

AB + CQ = AC + BQ

But AB = AC

Therefore, CQ = BQ or BQ = CQ