##### Class 10^{th} Mathematics Gujarat Board Solution

**Exercise 11.1**- A and B are the points on ⨀(O, r). bar ab is not a diameter of the circle.…
- A, B are the points on ⨀ (O, r) such that tangents at A and B intersect in P.…
- A, B are the points on ⨀(O, r) such that tangents at A and B to the circle…
- ⨀(O, r1) and ⨀(O, r2) are such that r1 r2. Chord AB of ⨀(O, r1) touches ⨀(O,…
- In example 4, if r1 = 41 and r2 = 9, find AB.

**Exercise 11.2**- P is the point in the exterior of ⨀(O, r) and the tangents from P to the circle…
- Two concentric circles having radii 73 and 55 are given. The chord of the…
- bar ab is a diameter of ⨀(O, 10). A tangent is drawn from B to ⨀(O, 8) which…
- P is in the exterior of a circle at distance 34 from the centre 0. A line…
- In figure 11.24, two tangents are drawn to a circle from a point A which is in…
- Prove that the perpendicular drawn to a tangent to the circle at the point of…
- Tangents from P, a point in the exterior of ⨀(O, r) touch the circle at A and…
- bar pt and bar pr are the tangents drawn to ⨀ (O, r) from point P lying in the…
- vector ab is a chord of ⨀(O, 5) such that AB = 8. Tangents at A and B to the…
- P lies in the exterior of ⨀(O, 5) such that OP = 13. Two tangents are drawn to…

**Exercise 11**- A circle touches the sides bar bc , bar ca , vector ab of ΔABC at points D, E,…
- ΔABC is an isosceles triangle in which vector ab congruent bar ac . A circle…
- ∠B is a right angle in ΔABC. If AB = 24, BC = 7, then find the radius of the…
- A circle touches all the three sides of a right angled ΔABC in which LB is…
- In â ABCD, m∠D = 90. A circle with centre 0 and radius r touches its sides bar…
- Two concentric circles are given. Prove that all chords of the circle with…
- A circle touches all the sides of â ABCD. If AB = 5, BC = 8, CD = 6. Find AD.…
- A circle touches all the sides of â ABCD. If vector ab is the largest side then…
- P is a point in the exterior of a circle having centre 0 and radius 24. OP =…
- P is in exterior of ⨀(O, 15). A tangent from P touches the circle at T. If PT…
- vector p_si , pb touch ⨀(O, 15) at A and B. If m∠AOB = 80, then m∠OPB =A. 80…
- A tangent from P, a point in the exterior of a circle, touches the circle at…
- In ΔABC, AB = 3, BC = 4, AC = 5, then the radius of the circle touching all…
- vector pq and vector pr touch the circle with centre 0 at A and B…
- The points of contact of the tangents from an exterior point P to the circle…
- A chord of ⨀(O, 5) touches ⨀(O, 3). Therefore the length of the chord =A. 8…

**Exercise 11.1**

- A and B are the points on ⨀(O, r). bar ab is not a diameter of the circle.…
- A, B are the points on ⨀ (O, r) such that tangents at A and B intersect in P.…
- A, B are the points on ⨀(O, r) such that tangents at A and B to the circle…
- ⨀(O, r1) and ⨀(O, r2) are such that r1 r2. Chord AB of ⨀(O, r1) touches ⨀(O,…
- In example 4, if r1 = 41 and r2 = 9, find AB.

**Exercise 11.2**

- P is the point in the exterior of ⨀(O, r) and the tangents from P to the circle…
- Two concentric circles having radii 73 and 55 are given. The chord of the…
- bar ab is a diameter of ⨀(O, 10). A tangent is drawn from B to ⨀(O, 8) which…
- P is in the exterior of a circle at distance 34 from the centre 0. A line…
- In figure 11.24, two tangents are drawn to a circle from a point A which is in…
- Prove that the perpendicular drawn to a tangent to the circle at the point of…
- Tangents from P, a point in the exterior of ⨀(O, r) touch the circle at A and…
- bar pt and bar pr are the tangents drawn to ⨀ (O, r) from point P lying in the…
- vector ab is a chord of ⨀(O, 5) such that AB = 8. Tangents at A and B to the…
- P lies in the exterior of ⨀(O, 5) such that OP = 13. Two tangents are drawn to…

**Exercise 11**

- A circle touches the sides bar bc , bar ca , vector ab of ΔABC at points D, E,…
- ΔABC is an isosceles triangle in which vector ab congruent bar ac . A circle…
- ∠B is a right angle in ΔABC. If AB = 24, BC = 7, then find the radius of the…
- A circle touches all the three sides of a right angled ΔABC in which LB is…
- In â ABCD, m∠D = 90. A circle with centre 0 and radius r touches its sides bar…
- Two concentric circles are given. Prove that all chords of the circle with…
- A circle touches all the sides of â ABCD. If AB = 5, BC = 8, CD = 6. Find AD.…
- A circle touches all the sides of â ABCD. If vector ab is the largest side then…
- P is a point in the exterior of a circle having centre 0 and radius 24. OP =…
- P is in exterior of ⨀(O, 15). A tangent from P touches the circle at T. If PT…
- vector p_si , pb touch ⨀(O, 15) at A and B. If m∠AOB = 80, then m∠OPB =A. 80…
- A tangent from P, a point in the exterior of a circle, touches the circle at…
- In ΔABC, AB = 3, BC = 4, AC = 5, then the radius of the circle touching all…
- vector pq and vector pr touch the circle with centre 0 at A and B…
- The points of contact of the tangents from an exterior point P to the circle…
- A chord of ⨀(O, 5) touches ⨀(O, 3). Therefore the length of the chord =A. 8…

###### Exercise 11.1

**Question 1.**A and B are the points on ⨀(O, r). is not a diameter of the circle. Prove that the tangents to the circle at A and B are not parallel.

**Answer:**Given that A and B are points on (O, r) and AB is a not a diameter of the circle.

We have to prove that the tangents to the circle at A and B are not parallel.

Proof:

Using the method of contradiction,

Let l and m be two parallel tangents to the circle have centre O drawn at the points A and B.

∴ OA ⊥ l and OB ⊥ m

Consider OA and OB perpendicular to l and m respectively and O is a common point.

∴ since l and m are two parallel lines, A – O – B

Hence, AB is a diameter, which contradicts with our assumption.

∴ Our assumption is wrong i.e. l and m are intersecting lines.

∴ Tangents to the circle at A and B are not parallel.

Hence proved.

**Question 2.**A, B are the points on ⨀ (O, r) such that tangents at A and B intersect in P. Prove that OP is the bisector of ∠AOB and PO is the bisector of ∠APB.

**Answer:**Given that A and B are the points on (O, r) such that tangents at A and B intersect in P.

We have to prove that OP is the bisector of ∠AOB and PO is the bisector of ∠APB.

Proof:

In circle (O, r), AP is a tangent at A and BP is the tangent at B.

⇒ ∠OAP = ∠OBP = 90°

Considering ΔOAP and ΔOBP,

⇒ OA = OB (radius)

⇒ OP = OP (common segment)

∴ By RHS theorem, ΔOAP = ΔOBP i.e. OAP and OBP is a congruence.

∴ ∠APO = ∠BOP and ∠AOP = ∠BOP

Here, O is in the interior part of ∠APB and P is in the interior part of ∠AOB.

∴ OP is the bisector of ∠AOB and PO is the bisector of ∠APB.

Hence proved.

**Question 3.**A, B are the points on ⨀(O, r) such that tangents at A and B to the circle intersect in P. Show that the circle with as a diameter passes through A and B.

**Answer:**Given that A and B are points on circle (O, r) such that tangents at A and B to the circle intersect in P.

We have to prove that the circle with OP as a diameter passes through A and B.

Proof:

__We know that a tangent drawn to a circle is perpendicular to the radius drawn from the point of contact.__

⇒ OA ⊥ AP and OB ⊥ PB

⇒ ∠OAP = 90° and ∠OBP = 90°

∴ ∠OAP + ∠OBP = 180° … (1)

For OAPB,

∠OAP + ∠APB + ∠AOB + ∠OBP = 360°

⇒ (∠APB + ∠AOB) + (∠OAP + ∠OBP) = 360°

From (1),

⇒ ∠APB + ∠AOB + 180° = 360°

∴ ∠APB + ∠AOB = 180° … (2)

From (1) and (2),

OAPB is cyclic.

Here, OP makes a right angle at A.

Then OP is a diameter.

∴ The circle with OP as a diameter passes through A and B.

Hence proved.

**Question 4.**⨀(O, r_{1}) and ⨀(O, r_{2}) are such that r_{1}> r_{2}. Chord AB of ⨀(O, r_{1}) touches ⨀(O, r_{2}). Find AB in terms of r_{1} and r_{2}.

**Answer:**

Given that circle (O, r_{1}) and circle (O, r_{2}) are such that r_{1} > r_{2}.

∴ The circles are concentric.

Let chord AB of circle (O, r_{1}) touches circle (O, r_{2}) at P.

Thus, AB is tangent to circle (O, r_{2}).

∴ OP ⊥ AB and P ∈ AB

Here, P is the foot of the perpendicular drawn from centre O on the chord AB of circle (O, r_{2}).

∴ P is the midpoint of AB.

⇒ AB = 2AP … (1)

Consider right angle ΔOPA,

By Pythagoras Theorem,

⇒ OA^{2} = AP^{2} + OP^{2}

⇒ r_{1}^{2} = AP^{2} + r_{2}^{2}

⇒ AP^{2} = r_{1}^{2} – r_{2}^{2}

∴ AP =

From (1),

∴ AB = 2

**Question 5.**In example 4, if r_{1} = 41 and r_{2} = 9, find AB.

**Answer:**From example 4,

Given that r_{1} = 41 and r_{2} = 9

Length of chord AB = 2

⇒ AB = 2

= 2

= 2

= 2 (40)

= 80

∴ AB = 80

**Question 1.**

A and B are the points on ⨀(O, r). is not a diameter of the circle. Prove that the tangents to the circle at A and B are not parallel.

**Answer:**

Given that A and B are points on (O, r) and AB is a not a diameter of the circle.

We have to prove that the tangents to the circle at A and B are not parallel.

Proof:

Using the method of contradiction,

Let l and m be two parallel tangents to the circle have centre O drawn at the points A and B.

∴ OA ⊥ l and OB ⊥ m

Consider OA and OB perpendicular to l and m respectively and O is a common point.

∴ since l and m are two parallel lines, A – O – B

Hence, AB is a diameter, which contradicts with our assumption.

∴ Our assumption is wrong i.e. l and m are intersecting lines.

∴ Tangents to the circle at A and B are not parallel.

Hence proved.

**Question 2.**

A, B are the points on ⨀ (O, r) such that tangents at A and B intersect in P. Prove that OP is the bisector of ∠AOB and PO is the bisector of ∠APB.

**Answer:**

Given that A and B are the points on (O, r) such that tangents at A and B intersect in P.

We have to prove that OP is the bisector of ∠AOB and PO is the bisector of ∠APB.

Proof:

In circle (O, r), AP is a tangent at A and BP is the tangent at B.

⇒ ∠OAP = ∠OBP = 90°

Considering ΔOAP and ΔOBP,

⇒ OA = OB (radius)

⇒ OP = OP (common segment)

∴ By RHS theorem, ΔOAP = ΔOBP i.e. OAP and OBP is a congruence.

∴ ∠APO = ∠BOP and ∠AOP = ∠BOP

Here, O is in the interior part of ∠APB and P is in the interior part of ∠AOB.

∴ OP is the bisector of ∠AOB and PO is the bisector of ∠APB.

Hence proved.

**Question 3.**

A, B are the points on ⨀(O, r) such that tangents at A and B to the circle intersect in P. Show that the circle with as a diameter passes through A and B.

**Answer:**

Given that A and B are points on circle (O, r) such that tangents at A and B to the circle intersect in P.

We have to prove that the circle with OP as a diameter passes through A and B.

Proof:

__We know that a tangent drawn to a circle is perpendicular to the radius drawn from the point of contact.__

⇒ OA ⊥ AP and OB ⊥ PB

⇒ ∠OAP = 90° and ∠OBP = 90°

∴ ∠OAP + ∠OBP = 180° … (1)

For OAPB,

∠OAP + ∠APB + ∠AOB + ∠OBP = 360°

⇒ (∠APB + ∠AOB) + (∠OAP + ∠OBP) = 360°

From (1),

⇒ ∠APB + ∠AOB + 180° = 360°

∴ ∠APB + ∠AOB = 180° … (2)

From (1) and (2),

OAPB is cyclic.

Here, OP makes a right angle at A.

Then OP is a diameter.

∴ The circle with OP as a diameter passes through A and B.

Hence proved.

**Question 4.**

⨀(O, r_{1}) and ⨀(O, r_{2}) are such that r_{1}> r_{2}. Chord AB of ⨀(O, r_{1}) touches ⨀(O, r_{2}). Find AB in terms of r_{1} and r_{2}.

**Answer:**

Given that circle (O, r_{1}) and circle (O, r_{2}) are such that r_{1} > r_{2}.

∴ The circles are concentric.

Let chord AB of circle (O, r_{1}) touches circle (O, r_{2}) at P.

Thus, AB is tangent to circle (O, r_{2}).

∴ OP ⊥ AB and P ∈ AB

Here, P is the foot of the perpendicular drawn from centre O on the chord AB of circle (O, r_{2}).

∴ P is the midpoint of AB.

⇒ AB = 2AP … (1)

Consider right angle ΔOPA,

By Pythagoras Theorem,

⇒ OA^{2} = AP^{2} + OP^{2}

⇒ r_{1}^{2} = AP^{2} + r_{2}^{2}

⇒ AP^{2} = r_{1}^{2} – r_{2}^{2}

∴ AP =

From (1),

∴ AB = 2

**Question 5.**

In example 4, if r_{1} = 41 and r_{2} = 9, find AB.

**Answer:**

From example 4,

Given that r_{1} = 41 and r_{2} = 9

Length of chord AB = 2

⇒ AB = 2

= 2

= 2

= 2 (40)

= 80

∴ AB = 80

###### Exercise 11.2

**Question 1.**P is the point in the exterior of ⨀(O, r) and the tangents from P to the circle touch the circle at X and Y.

(1) Find OP, if r = 12, XP = 5

(2) Find m∠XPO, if m∠XOY = 110

(3) Find r, if OP = 25 and PY = 24

(4) Find m∠XOP, if m∠XPO = 80

**Answer:**Given that P is the point in exterior of circle (O, r) and the tangents from P to the circle touch the circle at X and Y.

(1) Here OX is the radius of the circle and PX is the tangent.

∴ OX ⊥ PX

Given r = 12 = OX and XP = 5

For right angled ΔPXO,

⇒ OP^{2} = PX^{2} + OX^{2}

= 5^{2} + 12^{2}

= 25 + 144

= 169

⇒ OP^{2} = 169

∴ OP = 13

(2) Given ∠XOY = 110°

Here ∠XPY and ∠XOY are supplementary angles.

__We know that two angles are supplementary when they add upto 180°.__

⇒ ∠XPY + ∠XOY = 180°

⇒ ∠XPY + 110° = 180°

⇒ ∠XPY = 70°

Also, OP is a bisector pf ∠XPY

∴ ∠XPO = 1/2 ∠XPY = 1/2 (70°) = 35°

(3) OY is the radius of the circle and PY is a tangent.

∴ OY ⊥ PY

Now, OY = r,

Given that OP = 25 and PY = 24

In right angled ΔPYO,

By Pythagoras Theorem,

⇒ OP^{2} = OY^{2} + PY^{2}

⇒ 25^{2} = r^{2} + 24^{2}

⇒ r^{2} = 625 – 576

⇒ r^{2} = 49

∴ r = 7

(4) Given ∠XPO = 80°

Here, OX is the radius of the circle and PX is a tangent.

∴ OX ⊥ PX

Also, ∠XOP and ∠XPO are complementary angles.

We know that two angles are complementary when they add up to 90°.

⇒ ∠XOP + ∠XPO = 90°

⇒ ∠XOP + 80° = 90°

∴ ∠XOP = 10°

**Question 2.**Two concentric circles having radii 73 and 55 are given. The chord of the circle with larger radius touches the circle with smaller radius. Find the length of the chord.

**Answer:**Given the radius of larger circle = 73 = OB and radius of smaller circle = 55 = OM

Since AB is a tangent, OM ⊥ AB.

Consider ΔOMB,

Here, ∠OMB is a right angle.

By Pythagoras Theorem,

⇒ OB^{2} = OM^{2} + MB^{2}

⇒ MB^{2} = OB^{2} – OM^{2}

= 73^{2} – 55^{2}

We know that a^{2} – b^{2} = (a + b) (a – b)

⇒ MB^{2} = (73 + 55) (73 – 55)

= (128) (18)

= 2304

∴ MB = 48

Now, length of chord = AB = 2MB = 2 (48) = 96

∴ The length of chord is 96.

**Question 3.**is a diameter of ⨀(O, 10). A tangent is drawn from B to ⨀(O, 8) which touches ⨀(O, 8) at D. intersects 0(0, 10) in C. Find AC.

**Answer:**

Given that AB is a diameter of circle (O, 10).

⇒ OA = OB = 10 = radius

⇒ AB = 20 = diameter

Also given a tangent is drawn from B to circle (O, 8) which touches the circle at D.

⇒ OD = 8 = radius

Since BD is a tangent, OD ⊥ BD.

And since the angle is inscribed in a semi circle, ∠ACB = 90°.

⇒ ∠ODB = ∠ACB = 90°

∴ ∠DBO ≅ ∠CBA

By AA theorem,

Thus, correspondence ODB ↔ ACB is a similarity.

Then =

⇒ =

⇒ AC = = 16

∴ AC = 16

**Question 4.**P is in the exterior of a circle at distance 34 from the centre 0. A line through P touches the circle at Q. PQ = 16, find the diameter of the circle.

**Answer:**Given OP = 34, PQ = 16

OQ is the radius of the circle.

Since PQ is a tangent to the circle, OQ ⊥ PQ.

In right angled ΔOQP,

By Pythagoras Theorem,

⇒ OP^{2} = PQ^{2} + OQ^{2}

⇒ 34^{2} = 16^{2} + OQ^{2}

⇒ OQ^{2} = 1156 – 256

⇒ OQ^{2} = 900

⇒ OQ = 30

∴ Diameter of circle = 2r = 2 × OQ = 2 × 30 = 60

∴ Diameter = 60

**Question 5.**In figure 11.24, two tangents are drawn to a circle from a point A which is in the exterior of the circle. The points of contact of the tangents are P and Q as shown in the figure. A line 1 touches the circle at R and intersects and in B and C respectively. If AB = c, BC = a, CA = b, then prove that

(1) AP + AQ = a + b + c

(2) AB + BR = AC + CR = AP = AQ =

**Answer:**Given that AP and AQ are tangents to the circle.

A line l touches the circle at R and intersects AP and AQ in B and C respectively. And AB = c, BC = a, CA = b.

We have to prove that

(1) AP + AQ = a + b + c

(2) AB + BR = AC + CR = AP = AQ =

Proof:

By theorem,

AP = AQ, BP = BR and CQ = CR … (1)

(1) AP + AQ = (AB + BP) + (AC + CQ)

= (AB + BR) + (AC + CR) [From (1)]

= AB + AC + (BR + CR)

Since B – R – C,

⇒ AP + AQ = AB + AC + BC

= c + b + a

∴ AP + AQ = a + b + c … (2)

(2) AB + BR = AB + BP [From (1)]

= AP [∵ A – B – P]

= AQ [From (1)]

= AC + CQ [∵ A – C – Q]

= AC + CR [From (1)]

∴ AB + BR = AC + CR = AP = AQ … (3)

From (2),

⇒ AP + AQ = a + b + c

From (1),

⇒ AQ + AQ = a + b + c

⇒ 2AQ = a + b + c

⇒ AQ =

From (3),

∴ AB + BR = AC + CR = AP = AQ =

Hence proved.

**Question 6.**Prove that the perpendicular drawn to a tangent to the circle at the point of contact of the tangent passes through the centre of the circle.

**Answer:**Let l be a tangent to the circle having centre O. l touches the circle at P. Let m be the perpendicular line to l from P.

We have to prove that m passes through O i.e. O ∈ m.

Proof:

If O m, then we can find such M m that O and M are in the same half plane of l.

T ∈ l is a distinct point from P.

∴ ∠MPT = 90° and ∠OPT = 90°

M and O are points of the same half plane so this is impossible.

Thus, our assumption is wrong.

∴ O ∈ m

∴The perpendicular drawn to a tangent to the circle at the point of contact of the tangent passes through the centre of the circle.

Hence proved.

**Question 7.**Tangents from P, a point in the exterior of ⨀(O, r) touch the circle at A and B. Prove that and bisects.

**Answer:**Let PA and PB be tangents to the circle drawn from point P which is in the exterior of circle (O, r). Given A and B are on the circle.

We have to prove that OP ⊥ AB and OP bisects AB.

Proof:

Here PA and PB are tangents drawn to the circle from an exterior point P.

∴ OP intersects AB at C.

Also given that A and B are on the circle.

__We know that the tangents drawn to a circle from a point in the exterior of the circle are congruent.__

∴ PA = PB

⇒ OP = OP [common]

⇒ OA = OB [radii of circle]

∴ By SSS theorem,

ΔOAP ≅ ΔOBP

Then, ∠AOP = ∠BOP

⇒ ∠AOC = ∠BOC [C ∈ OP]

⇒ OA = OB [radii of circle]

⇒ OC = OC [common]

∴ By SAS theorem,

ΔAOC ≅ ΔBOC

∴ AC = BC and ∠ACO = ∠BCO = 90°

Now, as C ∈ OP,

⇒ OP bisects AB.

Also AC ⊥ OC and BC ⊥ OC

⇒ OP ⊥ AB [∵ A – C – B]

∴ OP ⊥ AB and OP bisects AB.

Hence proved.

**Question 8.** and are the tangents drawn to ⨀ (O, r) from point P lying in the exterior of the circle and T and R are their points of contact respectively. Prove that m∠TPR = 2m∠OTR.

**Answer:**Given that PT and PR are tangents drawn to circle (O, r) from point P lying in the exterior of the circle and T and R are their points of contact respectively.

We have to prove that m∠TPR = 2m∠OTR

Proof:

By theorem,

PT ≅ PR

__We know that angles opposite to congruent sides are equal.__

∴ ∠PTR = ∠PRT

We know that sum of all angles in a triangle is 180°.

In ΔPTR,

⇒ ∠PTR + ∠PRT + ∠TPR = 180°

⇒ ∠PTR + ∠PTR + ∠TPR = 180°

⇒ 2∠PTR + ∠TPR = 180°

⇒ 2∠PTR = 180° – ∠TPR

⇒ ∠PTR = 90° – 1/2 ∠TPR

∴ 1/2 ∠TPR = 90° – ∠PTR … (1)

Then, OT ⊥ PT,

⇒ ∠OTP = 90°

⇒ ∠OTR + ∠PTR = 90°

⇒ ∠OTR = 90° – ∠PTR … (2)

From (1) and (2),

⇒ 1/2 ∠TPR = ∠OTR

⇒ ∠TPR = 2∠OTR

∴ m∠TPR = 2m∠OTR

Hence proved.

**Question 9.**is a chord of ⨀(O, 5) such that AB = 8. Tangents at A and B to the circle intersect in P. Find PA.

**Answer:**Given AB is a chord of circle (O, 5) such that AB = 8.

Let PR = x.

Since OP is a perpendicular bisector of AB,

AR = BR = 4

Consider ΔORA where ∠R = 90°,

By Pythagoras Theorem,

⇒ OA^{2} = OR^{2} + AR^{2}

⇒ 5^{2} = OR^{2} + 4^{2}

⇒ OR^{2} = 25 – 16 = 9

∴ OR = 3

⇒ OP = PR + RO = x + 3

Consider ΔARP where ∠R = 90°,

⇒ PA^{2} = AR^{2} + PR^{2}

⇒ PA^{2} = 4^{2} + x^{2} = 16 + x^{2} … (1)

Consider ΔOAP where ∠A = 90°,

⇒ PA^{2} = OP^{2} – OA^{2}

= (x + 3)^{2} – (5)^{2}

= x^{2} + 9 + 6x – 25

= x^{2} + 6x – 16 … (2)

From (1) and (2),

⇒ 16 + x^{2} = x^{2} + 6x – 16

⇒ 6x = 32

⇒ x =

From (1),

⇒ PA^{2} = 16 + x^{2}

= 16 + () ^{2}

= 16 +

=

=

∴ PA =

**Question 10.**P lies in the exterior of ⨀(O, 5) such that OP = 13. Two tangents are drawn to the circle which touch the circle in A and B. Find AB.

**Answer:**Given P lies in the exterior of circle (O, 5) such that OP = 13.

Let OR = x.

Consider ΔOAP where ∠A = 90°,

By Pythagoras Theorem,

⇒ OP^{2} = OA^{2} + AP^{2}

⇒ 13^{2} = 5^{2} + AP^{2}

⇒ AP^{2} = 169 – 25 = 144

∴ AP = 12

Consider ΔORA where ∠R = 90°,

⇒ AR^{2} = OA^{2} – OR^{2}

⇒ AR^{2} = 5^{2} – x^{2} = 25 – x^{2} … (1)

Consider ΔARP where ∠R = 90°,

⇒ AR^{2} = AP^{2} – PR^{2}

= (12)^{2} – (13 – x)^{2}

= 144 – (13 + x^{2} – 26x)

= – x^{2} + 26x + 131 … (2)

From (1) and (2),

⇒ 25 – x^{2} = – x^{2} + 26x + 131

⇒ 26x = 50

⇒ x =

From (1),

⇒ AR^{2} = 25 – x^{2}

= 25 – () ^{2}

= 25 –

=

=

∴ AR =

But AB = 2AR,

⇒ AB = 2 ()

∴ AB = = 9.23

**Question 1.**

P is the point in the exterior of ⨀(O, r) and the tangents from P to the circle touch the circle at X and Y.

(1) Find OP, if r = 12, XP = 5

(2) Find m∠XPO, if m∠XOY = 110

(3) Find r, if OP = 25 and PY = 24

(4) Find m∠XOP, if m∠XPO = 80

**Answer:**

Given that P is the point in exterior of circle (O, r) and the tangents from P to the circle touch the circle at X and Y.

(1) Here OX is the radius of the circle and PX is the tangent.

∴ OX ⊥ PX

Given r = 12 = OX and XP = 5

For right angled ΔPXO,

⇒ OP^{2} = PX^{2} + OX^{2}

= 5^{2} + 12^{2}

= 25 + 144

= 169

⇒ OP^{2} = 169

∴ OP = 13

(2) Given ∠XOY = 110°

Here ∠XPY and ∠XOY are supplementary angles.

__We know that two angles are supplementary when they add upto 180°.__

⇒ ∠XPY + ∠XOY = 180°

⇒ ∠XPY + 110° = 180°

⇒ ∠XPY = 70°

Also, OP is a bisector pf ∠XPY

∴ ∠XPO = 1/2 ∠XPY = 1/2 (70°) = 35°

(3) OY is the radius of the circle and PY is a tangent.

∴ OY ⊥ PY

Now, OY = r,

Given that OP = 25 and PY = 24

In right angled ΔPYO,

By Pythagoras Theorem,

⇒ OP^{2} = OY^{2} + PY^{2}

⇒ 25^{2} = r^{2} + 24^{2}

⇒ r^{2} = 625 – 576

⇒ r^{2} = 49

∴ r = 7

(4) Given ∠XPO = 80°

Here, OX is the radius of the circle and PX is a tangent.

∴ OX ⊥ PX

Also, ∠XOP and ∠XPO are complementary angles.

We know that two angles are complementary when they add up to 90°.

⇒ ∠XOP + ∠XPO = 90°

⇒ ∠XOP + 80° = 90°

∴ ∠XOP = 10°

**Question 2.**

Two concentric circles having radii 73 and 55 are given. The chord of the circle with larger radius touches the circle with smaller radius. Find the length of the chord.

**Answer:**

Given the radius of larger circle = 73 = OB and radius of smaller circle = 55 = OM

Since AB is a tangent, OM ⊥ AB.

Consider ΔOMB,

Here, ∠OMB is a right angle.

By Pythagoras Theorem,

⇒ OB^{2} = OM^{2} + MB^{2}

⇒ MB^{2} = OB^{2} – OM^{2}

= 73^{2} – 55^{2}

We know that a^{2} – b^{2} = (a + b) (a – b)

⇒ MB^{2} = (73 + 55) (73 – 55)

= (128) (18)

= 2304

∴ MB = 48

Now, length of chord = AB = 2MB = 2 (48) = 96

∴ The length of chord is 96.

**Question 3.**

is a diameter of ⨀(O, 10). A tangent is drawn from B to ⨀(O, 8) which touches ⨀(O, 8) at D. intersects 0(0, 10) in C. Find AC.

**Answer:**

Given that AB is a diameter of circle (O, 10).

⇒ OA = OB = 10 = radius

⇒ AB = 20 = diameter

Also given a tangent is drawn from B to circle (O, 8) which touches the circle at D.

⇒ OD = 8 = radius

Since BD is a tangent, OD ⊥ BD.

And since the angle is inscribed in a semi circle, ∠ACB = 90°.

⇒ ∠ODB = ∠ACB = 90°

∴ ∠DBO ≅ ∠CBA

By AA theorem,

Thus, correspondence ODB ↔ ACB is a similarity.

Then =

⇒ =

⇒ AC = = 16

∴ AC = 16

**Question 4.**

P is in the exterior of a circle at distance 34 from the centre 0. A line through P touches the circle at Q. PQ = 16, find the diameter of the circle.

**Answer:**

Given OP = 34, PQ = 16

OQ is the radius of the circle.

Since PQ is a tangent to the circle, OQ ⊥ PQ.

In right angled ΔOQP,

By Pythagoras Theorem,

⇒ OP^{2} = PQ^{2} + OQ^{2}

⇒ 34^{2} = 16^{2} + OQ^{2}

⇒ OQ^{2} = 1156 – 256

⇒ OQ^{2} = 900

⇒ OQ = 30

∴ Diameter of circle = 2r = 2 × OQ = 2 × 30 = 60

∴ Diameter = 60

**Question 5.**

In figure 11.24, two tangents are drawn to a circle from a point A which is in the exterior of the circle. The points of contact of the tangents are P and Q as shown in the figure. A line 1 touches the circle at R and intersects and in B and C respectively. If AB = c, BC = a, CA = b, then prove that

(1) AP + AQ = a + b + c

(2) AB + BR = AC + CR = AP = AQ =

**Answer:**

Given that AP and AQ are tangents to the circle.

A line l touches the circle at R and intersects AP and AQ in B and C respectively. And AB = c, BC = a, CA = b.

We have to prove that

(1) AP + AQ = a + b + c

(2) AB + BR = AC + CR = AP = AQ =

Proof:

By theorem,

AP = AQ, BP = BR and CQ = CR … (1)

(1) AP + AQ = (AB + BP) + (AC + CQ)

= (AB + BR) + (AC + CR) [From (1)]

= AB + AC + (BR + CR)

Since B – R – C,

⇒ AP + AQ = AB + AC + BC

= c + b + a

∴ AP + AQ = a + b + c … (2)

(2) AB + BR = AB + BP [From (1)]

= AP [∵ A – B – P]

= AQ [From (1)]

= AC + CQ [∵ A – C – Q]

= AC + CR [From (1)]

∴ AB + BR = AC + CR = AP = AQ … (3)

From (2),

⇒ AP + AQ = a + b + c

From (1),

⇒ AQ + AQ = a + b + c

⇒ 2AQ = a + b + c

⇒ AQ =

From (3),

∴ AB + BR = AC + CR = AP = AQ =

Hence proved.

**Question 6.**

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

**Answer:**

Let l be a tangent to the circle having centre O. l touches the circle at P. Let m be the perpendicular line to l from P.

We have to prove that m passes through O i.e. O ∈ m.

Proof:

If O m, then we can find such M m that O and M are in the same half plane of l.

T ∈ l is a distinct point from P.

∴ ∠MPT = 90° and ∠OPT = 90°

M and O are points of the same half plane so this is impossible.

Thus, our assumption is wrong.

∴ O ∈ m

∴The perpendicular drawn to a tangent to the circle at the point of contact of the tangent passes through the centre of the circle.

Hence proved.

**Question 7.**

Tangents from P, a point in the exterior of ⨀(O, r) touch the circle at A and B. Prove that and bisects.

**Answer:**

Let PA and PB be tangents to the circle drawn from point P which is in the exterior of circle (O, r). Given A and B are on the circle.

We have to prove that OP ⊥ AB and OP bisects AB.

Proof:

Here PA and PB are tangents drawn to the circle from an exterior point P.

∴ OP intersects AB at C.

Also given that A and B are on the circle.

__We know that the tangents drawn to a circle from a point in the exterior of the circle are congruent.__

∴ PA = PB

⇒ OP = OP [common]

⇒ OA = OB [radii of circle]

∴ By SSS theorem,

ΔOAP ≅ ΔOBP

Then, ∠AOP = ∠BOP

⇒ ∠AOC = ∠BOC [C ∈ OP]

⇒ OA = OB [radii of circle]

⇒ OC = OC [common]

∴ By SAS theorem,

ΔAOC ≅ ΔBOC

∴ AC = BC and ∠ACO = ∠BCO = 90°

Now, as C ∈ OP,

⇒ OP bisects AB.

Also AC ⊥ OC and BC ⊥ OC

⇒ OP ⊥ AB [∵ A – C – B]

∴ OP ⊥ AB and OP bisects AB.

Hence proved.

**Question 8.**

and are the tangents drawn to ⨀ (O, r) from point P lying in the exterior of the circle and T and R are their points of contact respectively. Prove that m∠TPR = 2m∠OTR.

**Answer:**

Given that PT and PR are tangents drawn to circle (O, r) from point P lying in the exterior of the circle and T and R are their points of contact respectively.

We have to prove that m∠TPR = 2m∠OTR

Proof:

By theorem,

PT ≅ PR

__We know that angles opposite to congruent sides are equal.__

∴ ∠PTR = ∠PRT

We know that sum of all angles in a triangle is 180°.

In ΔPTR,

⇒ ∠PTR + ∠PRT + ∠TPR = 180°

⇒ ∠PTR + ∠PTR + ∠TPR = 180°

⇒ 2∠PTR + ∠TPR = 180°

⇒ 2∠PTR = 180° – ∠TPR

⇒ ∠PTR = 90° – 1/2 ∠TPR

∴ 1/2 ∠TPR = 90° – ∠PTR … (1)

Then, OT ⊥ PT,

⇒ ∠OTP = 90°

⇒ ∠OTR + ∠PTR = 90°

⇒ ∠OTR = 90° – ∠PTR … (2)

From (1) and (2),

⇒ 1/2 ∠TPR = ∠OTR

⇒ ∠TPR = 2∠OTR

∴ m∠TPR = 2m∠OTR

Hence proved.

**Question 9.**

is a chord of ⨀(O, 5) such that AB = 8. Tangents at A and B to the circle intersect in P. Find PA.

**Answer:**

Given AB is a chord of circle (O, 5) such that AB = 8.

Let PR = x.

Since OP is a perpendicular bisector of AB,

AR = BR = 4

Consider ΔORA where ∠R = 90°,

By Pythagoras Theorem,

⇒ OA^{2} = OR^{2} + AR^{2}

⇒ 5^{2} = OR^{2} + 4^{2}

⇒ OR^{2} = 25 – 16 = 9

∴ OR = 3

⇒ OP = PR + RO = x + 3

Consider ΔARP where ∠R = 90°,

⇒ PA^{2} = AR^{2} + PR^{2}

⇒ PA^{2} = 4^{2} + x^{2} = 16 + x^{2} … (1)

Consider ΔOAP where ∠A = 90°,

⇒ PA^{2} = OP^{2} – OA^{2}

= (x + 3)^{2} – (5)^{2}

= x^{2} + 9 + 6x – 25

= x^{2} + 6x – 16 … (2)

From (1) and (2),

⇒ 16 + x^{2} = x^{2} + 6x – 16

⇒ 6x = 32

⇒ x =

From (1),

⇒ PA^{2} = 16 + x^{2}

= 16 + () ^{2}

= 16 +

=

=

∴ PA =

**Question 10.**

P lies in the exterior of ⨀(O, 5) such that OP = 13. Two tangents are drawn to the circle which touch the circle in A and B. Find AB.

**Answer:**

Given P lies in the exterior of circle (O, 5) such that OP = 13.

Let OR = x.

Consider ΔOAP where ∠A = 90°,

By Pythagoras Theorem,

⇒ OP^{2} = OA^{2} + AP^{2}

⇒ 13^{2} = 5^{2} + AP^{2}

⇒ AP^{2} = 169 – 25 = 144

∴ AP = 12

Consider ΔORA where ∠R = 90°,

⇒ AR^{2} = OA^{2} – OR^{2}

⇒ AR^{2} = 5^{2} – x^{2} = 25 – x^{2} … (1)

Consider ΔARP where ∠R = 90°,

⇒ AR^{2} = AP^{2} – PR^{2}

= (12)^{2} – (13 – x)^{2}

= 144 – (13 + x^{2} – 26x)

= – x^{2} + 26x + 131 … (2)

From (1) and (2),

⇒ 25 – x^{2} = – x^{2} + 26x + 131

⇒ 26x = 50

⇒ x =

From (1),

⇒ AR^{2} = 25 – x^{2}

= 25 – () ^{2}

= 25 –

=

=

∴ AR =

But AB = 2AR,

⇒ AB = 2 ()

∴ AB = = 9.23

###### Exercise 11

**Question 1.**A circle touches the sides , , of ΔABC at points D, E, F respectively. BD = x, CE = y, AF = z. Prove that the area of ΔABC = .

**Answer:**

Given that a circle touches the sides BC, CA and AB of ΔABC at points D, E and F respectively.

Let BD = x, CE = y and AF = z.

We have to prove that area of ΔABC =

Proof:

BD and BF are tangents drawn from B. And D and F are points of contact.

∴ BD = BF = x

Similarly for tangents CE and CD drawn by C and tangents AE and AF drawn from A,

CE = CD = y and AF = AE = z

Sides of ΔABC,

⇒ AB = c = AF + BF = z + x … (1)

⇒ BC = a = BD + DC = x + y … (2)

⇒ CA = b = CE + AE = y + z … (3)

In ΔABC, 2s = AB + BC + AC

= (z + x) + (x + y) + (y + z)

= 2 (x + y + z)

∴ s = x + y + z … (4)

We know that area of ΔABC =

Area =

=

=

Hence proved.

**Question 2.**ΔABC is an isosceles triangle in which. A circle touching all the three sides of ΔABC touches at D. Prove that D is the mid – point of.

**Answer:**Given that ΔABC is an isosceles triangle in which AB ≅ AC and a circle touching all the three sides of ΔABC touches BC at D.

We have to prove that D is the midpoint of BC.

Proof:

__We know that if a circle touches all sides of a triangle, the lengths of tangents drawn from each vertex are equal.__

∴ AE = AF, BD = BF and CD = CE … (1)

Consider AB = AC,

Subtracting AF from both sides,

⇒ AB – AF = AC – AF

From (1),

⇒ AB – AF = AC – AE

Since A – F – B and A – E – C,

⇒ BF = CE

From (1),

⇒ BD = CD

Since B – D – C and BD = CD,

D is the midpoint of BC.

**Question 3.**∠B is a right angle in ΔABC. If AB = 24, BC = 7, then find the radius of the circle which touches all the three sides of ΔABC.

**Answer:**

Let radius be r and the centre of the circle touching all the sides of a triangle be I i.e. incentre.

In the figure,

ID = IE = IF = r

Since ΔABC is a right angled triangle, ∠B = 90°.

Also ID ⊥ BC and AB ⊥ BC.

∴ ID || AB and ID || FB

Similarly, IF || BD

∴ IFBD is a parallelogram.

∴ FB = ID = r and BD = IF = r … (1)

∴ Parallelogram IFBD is a rhombus.

Since ∠B = 90°, parallelogram IFBD is a square.

By Pythagoras Theorem,

⇒ AC^{2} = AB^{2} + BC^{2}

= 24^{2} + 7^{2}

= 576 + 49

= 625

∴ AC = 25

⇒ AB + BC + AC = 24 + 7 + 25

⇒ AF + FB + BD + DC + AC = 56

⇒ AE + r + r + CE + AC = 56

⇒ 2r + (AE + CE) + AC = 56

⇒ 2r + 2AC = 56

⇒ 2r + 2(25) = 56

⇒ r + 25 = 28

⇒ r = 3

∴ The radius of circle is 3.

**Question 4.**A circle touches all the three sides of a right angled ΔABC in which LB is right angle. Prove that the radius of the circle is

**Answer:**

Given that a circle touches all the three sides of a right angled ΔABC.

We have to prove that radius of a circle =

Proof:

Let radius be r and the centre of the circle touching all the sides of a triangle be I i.e. incentre.

In the figure,

ID = IE = IF = r

Since ΔABC is a right angled triangle, ∠B = 90°.

Also ID ⊥ BC and AB ⊥ BC.

∴ ID || AB and ID || FB

Similarly, IF || BD

∴ IFBD is a parallelogram.

∴ FB = ID = r and BD = IF = r … (1)

∴ Parallelogram IFBD is a rhombus.

Since ∠B = 90°, parallelogram IFBD is a square.

Now AE = AF

⇒ AE = AB – FB

= AB – r [From (1)] … (2)

And CE = CD

⇒ CE = BC – BD

= BC – r [From (1)] … (3)

Now, AC = AE + CE,

⇒ AC = AB – r + BC – r

⇒ AC = AB + BC – 2r

⇒ 2r = AB + BC – AC

⇒ r =

∴ The radius of a circle is .

**Question 5.**In â ABCD, m∠D = 90. A circle with centre 0 and radius r touches its sides and in P, Q, R and S respectively. If BC = 40, CD = 30 and BP = 25, then find the radius of the circle.

**Answer:**Given that in â ABCD, ∠D = 90°. BC = 40, CD = 30 and BP = 25

__We know that tangents drawn to a circle are perpendicular to the radius of the circle.__

⇒ ∠ORD = ∠OSD = 90°

Given ∠D = 90° and OR = OS = radius.

∴ ORDS is a square.

__We know that the tangents drawn to a circle from a point in the exterior of the circle are congruent.__

∴ BP = BQ, CQ = CR and DR = DS.

Consider BP = BQ,

⇒ BQ = 25 [BP = 25]

⇒ BC – CQ = 40

⇒ CQ = 40 – 25 = 15 [BC = 40]

Consider CQ = CR,

⇒ CR = 15

⇒ CD – DR = 15

⇒ DR = 30 – 15 = 15 [CD = 30]

But ORDS is a square.

∴ OR = DR = 15

∴ Radius of circle OR is 15.

**Question 6.**Two concentric circles are given. Prove that all chords of the circle with larger radius which touch the circle with smaller radius are congruent.

**Answer:**Given are two concentric circles.

Let two chords PQ and RS of the circle with larger radius touch the circle with smaller radius.

We have to prove that PQ ≅ RS.

Proof:

Let two chords PQ and RS of the circle with larger radius touch the circle with smaller radius at points M and N respectively.

PQ and RS are tangents to the circle with smaller radius,

∴ OM = ON = radius of smaller circle.

Thus, chords AB and CD are equidistant from the centre of the circle with larger radius.

∴ PQ = RS

∴ PQ ≅ RS

∴All chords of the circle with larger radius which touch the circle with smaller radius are congruent.

Hence proved.

**Question 7.**A circle touches all the sides of â ABCD. If AB = 5, BC = 8, CD = 6. Find AD.

**Answer:**

We know that if a circle touches all the sides of a quadrilateral, then AB + CD = BC + DA

Given AB = 5, BC = 8, CD = 6

⇒ 5 + 6 = 8 + DA

⇒ 11 = 8 + DA

⇒ DA = 3

∴ AD = 3

**Question 8.**A circle touches all the sides of â ABCD. If is the largest side then prove that is the smallest side.

**Answer:**

Given that a circle touches all the sides of ABCD and AB is the largest side.

We have to prove that CD is the smallest side.

Proof:

The circle touches all sides of ABCD.

∴ AB + CD = BC + DA … (1)

Given AB is the largest side.

⇒ AB > BC

∴ AB = BC + m

From (1),

⇒ BC + m + CD = BC + DA

⇒ CD + m = DA

∴ CD < DA

Hence CD is smaller than DA. … (2)

But AB is the largest side.

⇒ AB > DA

∴ AB = DA + n

From (1),

⇒ DA + n + CD = BC + DA

⇒ CD + n = BC

∴ CD < BC

Hence CD is smaller than BC. … (3)

AB is largest side, so CD is smaller than AB. … (4)

From (2), (3) and (4), CD is the smallest side of ABCD.

**Question 9.**P is a point in the exterior of a circle having centre 0 and radius 24. OP = 25. A tangent from P touches the circle at Q. Find PQ.

**Answer:**

Given P lies in the exterior of a circle having centre O and PQ is a tangent.

∴ OQ ⊥ PQ

Also given OP = 25 and OQ = 24.

Consider ΔOQP,

∠OQP = 90°

By Pythagoras Theorem,

⇒ OP^{2} = OQ^{2} + PQ^{2}

⇒ 25^{2} = 24^{2} + PQ^{2}

⇒ PQ^{2} = 625 – 576

= 49

∴ PQ = 7

**Question 10.**P is in exterior of ⨀(O, 15). A tangent from P touches the circle at T. If PT = 8, then OP = ………..

A. 17

B. 13

C. 23

D. 7

**Answer:**Given P is in exterior of circle (O, 15) and a tangent from P touches the circle at T.

Thus, ∠OTP = 90°

By Pythagoras Theorem,

⇒ OP^{2} = OT^{2} + TP^{2}

⇒ OP^{2} = 15^{2} + 8^{2}

= 225 + 64

= 289

∴ OP = 17

**Question 11.**, touch ⨀(O, 15) at A and B. If m∠AOB = 80, then m∠OPB =

A. 80

B. 50

C. 10

D. 100

**Answer:**Given PA and PB touch circle (O, 15) at A and B and m ∠AOB = 80°.

Here, ΔPOA and ΔPOB are congruent right angled triangle.

⇒ ∠BOP = 1/2 ∠AOB

= 1/2 × 80°

= 40°

In right angled ΔOBP,

We know that sum of angles in a triangle is 180°.

⇒ ∠BOP + ∠B + ∠OPB = 180°

⇒ 40 + 90 + ∠OPB = 180°

⇒ 130 + ∠OPB = 180°

∴ ∠OPB = 50°

**Question 12.**A tangent from P, a point in the exterior of a circle, touches the circle at Q. If OP = 13, PQ = 5, then the diameter of the circle is

A. 576

B. 15

C. 8

D. 24

**Answer:**Given OP = 13 and PQ = 5

In right angled ΔOQP,

By Pythagoras Theorem,

⇒ OP^{2} = OQ^{2} + PQ^{2}

⇒ 13^{2} = r^{2} + 5^{2}

⇒ r^{2} = 169 – 25 = 144

∴ r = 12

Diameter of circle = 2r

= 2 (12)

= 24

∴ Diameter = 24

**Question 13.**In ΔABC, AB = 3, BC = 4, AC = 5, then the radius of the circle touching all the three sides is ………..

A. 2

B. 1

C. 4

D. 3

**Answer:**Given in ΔABC, AB = 3, BC = 4, AC = 5.

By Pythagoras Theorem,

⇒ AC^{2} = AB^{2}+ BC^{2}

⇒ 5^{2} = 3^{2} + 4^{2}

⇒ 5^{2} = 9 + 16

⇒ 5^{2} = 5^{2}

∴ ΔABC is a right angled triangle and ∠B is a right angle.

We know that the radius of the circle touching all the sides is .

⇒ The required radius of circle =

=

= 1

**Question 14.**and touch the circle with centre 0 at A and B respectively. If m∠OPB = 30 and OP = 10, then radius of the circle =

A. 5

B. 20

C. 60

D. 10

**Answer:**Given ∠OPB = 30° and OP = 10

In right angled ΔOBP,

Consider sin30° =

⇒ 1/2 =

⇒ OB = 5

∴ Radius of circle = OB = 5

**Question 15.**The points of contact of the tangents from an exterior point P to the circle with centre 0 are A and B. If m∠OPB = 30, then m∠ AOB = ……

A. 30

B. 60

C. 90

D. 120

**Answer:**In right angled ΔOBP,

Given ∠OBP = 30°

⇒ ∠BOP + ∠OPB + ∠B = 180°

⇒ ∠BOP + 30° + 90° = 180°

⇒ ∠BOP = 180° – 120° = 60°

∴ ∠BOP = 60°

Now, ∠AOB = 2 ∠BOP

= 2 (60°)

= 120°

∴ ∠AOB = 120°

**Question 16.**A chord of ⨀(O, 5) touches ⨀(O, 3). Therefore the length of the chord =

A. 8

B. 10

C. 7

D. 6

**Answer:**Given chord of circle (O, 5) touches circle (O, 3).

Radius of smaller circle OM = 3 and radius of bigger circle OB = 5

In right angled ΔOMB,

⇒ OB^{2} = OM^{2} + MB^{2}

⇒ 5^{2} = 3^{2} + MB^{2}

⇒ MB^{2} = 25 – 9 = 16

⇒ MB = 4

Length of chord AB = 2 (MB) = 2 (4) = 8

**Question 1.**

A circle touches the sides , , of ΔABC at points D, E, F respectively. BD = x, CE = y, AF = z. Prove that the area of ΔABC = .

**Answer:**

Given that a circle touches the sides BC, CA and AB of ΔABC at points D, E and F respectively.

Let BD = x, CE = y and AF = z.

We have to prove that area of ΔABC =

Proof:

BD and BF are tangents drawn from B. And D and F are points of contact.

∴ BD = BF = x

Similarly for tangents CE and CD drawn by C and tangents AE and AF drawn from A,

CE = CD = y and AF = AE = z

Sides of ΔABC,

⇒ AB = c = AF + BF = z + x … (1)

⇒ BC = a = BD + DC = x + y … (2)

⇒ CA = b = CE + AE = y + z … (3)

In ΔABC, 2s = AB + BC + AC

= (z + x) + (x + y) + (y + z)

= 2 (x + y + z)

∴ s = x + y + z … (4)

We know that area of ΔABC =

Area =

=

=

Hence proved.

**Question 2.**

ΔABC is an isosceles triangle in which. A circle touching all the three sides of ΔABC touches at D. Prove that D is the mid – point of.

**Answer:**

Given that ΔABC is an isosceles triangle in which AB ≅ AC and a circle touching all the three sides of ΔABC touches BC at D.

We have to prove that D is the midpoint of BC.

Proof:

__We know that if a circle touches all sides of a triangle, the lengths of tangents drawn from each vertex are equal.__

∴ AE = AF, BD = BF and CD = CE … (1)

Consider AB = AC,

Subtracting AF from both sides,

⇒ AB – AF = AC – AF

From (1),

⇒ AB – AF = AC – AE

Since A – F – B and A – E – C,

⇒ BF = CE

From (1),

⇒ BD = CD

Since B – D – C and BD = CD,

D is the midpoint of BC.

**Question 3.**

∠B is a right angle in ΔABC. If AB = 24, BC = 7, then find the radius of the circle which touches all the three sides of ΔABC.

**Answer:**

Let radius be r and the centre of the circle touching all the sides of a triangle be I i.e. incentre.

In the figure,

ID = IE = IF = r

Since ΔABC is a right angled triangle, ∠B = 90°.

Also ID ⊥ BC and AB ⊥ BC.

∴ ID || AB and ID || FB

Similarly, IF || BD

∴ IFBD is a parallelogram.

∴ FB = ID = r and BD = IF = r … (1)

∴ Parallelogram IFBD is a rhombus.

Since ∠B = 90°, parallelogram IFBD is a square.

By Pythagoras Theorem,

⇒ AC^{2} = AB^{2} + BC^{2}

= 24^{2} + 7^{2}

= 576 + 49

= 625

∴ AC = 25

⇒ AB + BC + AC = 24 + 7 + 25

⇒ AF + FB + BD + DC + AC = 56

⇒ AE + r + r + CE + AC = 56

⇒ 2r + (AE + CE) + AC = 56

⇒ 2r + 2AC = 56

⇒ 2r + 2(25) = 56

⇒ r + 25 = 28

⇒ r = 3

∴ The radius of circle is 3.

**Question 4.**

A circle touches all the three sides of a right angled ΔABC in which LB is right angle. Prove that the radius of the circle is

**Answer:**

Given that a circle touches all the three sides of a right angled ΔABC.

We have to prove that radius of a circle =

Proof:

Let radius be r and the centre of the circle touching all the sides of a triangle be I i.e. incentre.

In the figure,

ID = IE = IF = r

Since ΔABC is a right angled triangle, ∠B = 90°.

Also ID ⊥ BC and AB ⊥ BC.

∴ ID || AB and ID || FB

Similarly, IF || BD

∴ IFBD is a parallelogram.

∴ FB = ID = r and BD = IF = r … (1)

∴ Parallelogram IFBD is a rhombus.

Since ∠B = 90°, parallelogram IFBD is a square.

Now AE = AF

⇒ AE = AB – FB

= AB – r [From (1)] … (2)

And CE = CD

⇒ CE = BC – BD

= BC – r [From (1)] … (3)

Now, AC = AE + CE,

⇒ AC = AB – r + BC – r

⇒ AC = AB + BC – 2r

⇒ 2r = AB + BC – AC

⇒ r =

∴ The radius of a circle is .

**Question 5.**

In â ABCD, m∠D = 90. A circle with centre 0 and radius r touches its sides and in P, Q, R and S respectively. If BC = 40, CD = 30 and BP = 25, then find the radius of the circle.

**Answer:**

Given that in â ABCD, ∠D = 90°. BC = 40, CD = 30 and BP = 25

__We know that tangents drawn to a circle are perpendicular to the radius of the circle.__

⇒ ∠ORD = ∠OSD = 90°

Given ∠D = 90° and OR = OS = radius.

∴ ORDS is a square.

__We know that the tangents drawn to a circle from a point in the exterior of the circle are congruent.__

∴ BP = BQ, CQ = CR and DR = DS.

Consider BP = BQ,

⇒ BQ = 25 [BP = 25]

⇒ BC – CQ = 40

⇒ CQ = 40 – 25 = 15 [BC = 40]

Consider CQ = CR,

⇒ CR = 15

⇒ CD – DR = 15

⇒ DR = 30 – 15 = 15 [CD = 30]

But ORDS is a square.

∴ OR = DR = 15

∴ Radius of circle OR is 15.

**Question 6.**

Two concentric circles are given. Prove that all chords of the circle with larger radius which touch the circle with smaller radius are congruent.

**Answer:**

Given are two concentric circles.

Let two chords PQ and RS of the circle with larger radius touch the circle with smaller radius.

We have to prove that PQ ≅ RS.

Proof:

Let two chords PQ and RS of the circle with larger radius touch the circle with smaller radius at points M and N respectively.

PQ and RS are tangents to the circle with smaller radius,

∴ OM = ON = radius of smaller circle.

Thus, chords AB and CD are equidistant from the centre of the circle with larger radius.

∴ PQ = RS

∴ PQ ≅ RS

∴All chords of the circle with larger radius which touch the circle with smaller radius are congruent.

Hence proved.

**Question 7.**

A circle touches all the sides of â ABCD. If AB = 5, BC = 8, CD = 6. Find AD.

**Answer:**

We know that if a circle touches all the sides of a quadrilateral, then AB + CD = BC + DA

Given AB = 5, BC = 8, CD = 6

⇒ 5 + 6 = 8 + DA

⇒ 11 = 8 + DA

⇒ DA = 3

∴ AD = 3

**Question 8.**

A circle touches all the sides of â ABCD. If is the largest side then prove that is the smallest side.

**Answer:**

Given that a circle touches all the sides of ABCD and AB is the largest side.

We have to prove that CD is the smallest side.

Proof:

The circle touches all sides of ABCD.

∴ AB + CD = BC + DA … (1)

Given AB is the largest side.

⇒ AB > BC

∴ AB = BC + m

From (1),

⇒ BC + m + CD = BC + DA

⇒ CD + m = DA

∴ CD < DA

Hence CD is smaller than DA. … (2)

But AB is the largest side.

⇒ AB > DA

∴ AB = DA + n

From (1),

⇒ DA + n + CD = BC + DA

⇒ CD + n = BC

∴ CD < BC

Hence CD is smaller than BC. … (3)

AB is largest side, so CD is smaller than AB. … (4)

From (2), (3) and (4), CD is the smallest side of ABCD.

**Question 9.**

P is a point in the exterior of a circle having centre 0 and radius 24. OP = 25. A tangent from P touches the circle at Q. Find PQ.

**Answer:**

Given P lies in the exterior of a circle having centre O and PQ is a tangent.

∴ OQ ⊥ PQ

Also given OP = 25 and OQ = 24.

Consider ΔOQP,

∠OQP = 90°

By Pythagoras Theorem,

⇒ OP^{2} = OQ^{2} + PQ^{2}

⇒ 25^{2} = 24^{2} + PQ^{2}

⇒ PQ^{2} = 625 – 576

= 49

∴ PQ = 7

**Question 10.**

P is in exterior of ⨀(O, 15). A tangent from P touches the circle at T. If PT = 8, then OP = ………..

A. 17

B. 13

C. 23

D. 7

**Answer:**

Given P is in exterior of circle (O, 15) and a tangent from P touches the circle at T.

Thus, ∠OTP = 90°

By Pythagoras Theorem,

⇒ OP^{2} = OT^{2} + TP^{2}

⇒ OP^{2} = 15^{2} + 8^{2}

= 225 + 64

= 289

∴ OP = 17

**Question 11.**

, touch ⨀(O, 15) at A and B. If m∠AOB = 80, then m∠OPB =

A. 80

B. 50

C. 10

D. 100

**Answer:**

Given PA and PB touch circle (O, 15) at A and B and m ∠AOB = 80°.

Here, ΔPOA and ΔPOB are congruent right angled triangle.

⇒ ∠BOP = 1/2 ∠AOB

= 1/2 × 80°

= 40°

In right angled ΔOBP,

We know that sum of angles in a triangle is 180°.

⇒ ∠BOP + ∠B + ∠OPB = 180°

⇒ 40 + 90 + ∠OPB = 180°

⇒ 130 + ∠OPB = 180°

∴ ∠OPB = 50°

**Question 12.**

A tangent from P, a point in the exterior of a circle, touches the circle at Q. If OP = 13, PQ = 5, then the diameter of the circle is

A. 576

B. 15

C. 8

D. 24

**Answer:**

Given OP = 13 and PQ = 5

In right angled ΔOQP,

By Pythagoras Theorem,

⇒ OP^{2} = OQ^{2} + PQ^{2}

⇒ 13^{2} = r^{2} + 5^{2}

⇒ r^{2} = 169 – 25 = 144

∴ r = 12

Diameter of circle = 2r

= 2 (12)

= 24

∴ Diameter = 24

**Question 13.**

In ΔABC, AB = 3, BC = 4, AC = 5, then the radius of the circle touching all the three sides is ………..

A. 2

B. 1

C. 4

D. 3

**Answer:**

Given in ΔABC, AB = 3, BC = 4, AC = 5.

By Pythagoras Theorem,

⇒ AC^{2} = AB^{2}+ BC^{2}

⇒ 5^{2} = 3^{2} + 4^{2}

⇒ 5^{2} = 9 + 16

⇒ 5^{2} = 5^{2}

∴ ΔABC is a right angled triangle and ∠B is a right angle.

We know that the radius of the circle touching all the sides is .

⇒ The required radius of circle =

=

= 1

**Question 14.**

and touch the circle with centre 0 at A and B respectively. If m∠OPB = 30 and OP = 10, then radius of the circle =

A. 5

B. 20

C. 60

D. 10

**Answer:**

Given ∠OPB = 30° and OP = 10

In right angled ΔOBP,

Consider sin30° =

⇒ 1/2 =

⇒ OB = 5

∴ Radius of circle = OB = 5

**Question 15.**

The points of contact of the tangents from an exterior point P to the circle with centre 0 are A and B. If m∠OPB = 30, then m∠ AOB = ……

A. 30

B. 60

C. 90

D. 120

**Answer:**

In right angled ΔOBP,

Given ∠OBP = 30°

⇒ ∠BOP + ∠OPB + ∠B = 180°

⇒ ∠BOP + 30° + 90° = 180°

⇒ ∠BOP = 180° – 120° = 60°

∴ ∠BOP = 60°

Now, ∠AOB = 2 ∠BOP

= 2 (60°)

= 120°

∴ ∠AOB = 120°

**Question 16.**

A chord of ⨀(O, 5) touches ⨀(O, 3). Therefore the length of the chord =

A. 8

B. 10

C. 7

D. 6

**Answer:**

Given chord of circle (O, 5) touches circle (O, 3).

Radius of smaller circle OM = 3 and radius of bigger circle OB = 5

In right angled ΔOMB,

⇒ OB^{2} = OM^{2} + MB^{2}

⇒ 5^{2} = 3^{2} + MB^{2}

⇒ MB^{2} = 25 – 9 = 16

⇒ MB = 4

Length of chord AB = 2 (MB) = 2 (4) = 8