**Q. 1: In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30° , prove that BA : AT = 2 : 1.**

**Sol: **

AB is the chord passing through the centre

So, AB is the diameter

Since, angle in a semi circle is a right angle

∠APB= 90°

By using alternate segment theorem

We have ∠APB = ∠PAT = 30°

Now, in ⧍APB

∠BAP + ∠APB + ∠BAP = 180′ (Angle sum property of triangle)

∠BAP = 180° – 90° – 30° = 60°

Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)

60° = 30° + ∠PTA

∠PTA = 60° – 30° = 30°

We know that sides opposite to equal angles are equal

AP = AT

In right triangle ABP:

sin ABP=APBA

=> sin30° = ATBA =>12=ATBA

BA : AT = 2 : 1

**Q.2: In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6cm, BC = 9cm and CD = 8cm. Find the length of side AD.**

**Sol:**

We know that when a quadrilateral circumscribes a circle then the sum of opposite sides is equal to the sum of other opposite sides.

AB + CD = AD + BC

=> 6 + 8 = AD + 9

=> AD = 5 cm.

**Q.3: In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50° then what is the measure of ∠OAB.**

**Sol:**

We know that the radius and tangent are perpendicular at their point of contact

∠OBP = ∠OAF = 90°

Now, in quadrilateral AOBP:

∠A0B + ∠OBP + ∠APB + ∠OAP = 360°

∠A0B + 90° + 50°+ 90° = 360°

=> 230°+ ∠BOC = 360° ∠A0B = 130°

Now, in isosceles triangle AOB,

∠A0B + ∠OAB + ∠OBA = 180°

=> 130° + 2∠OAB = 180°

Therefore, ∠OAB = 25°

**Q.4: In the given figure, O is the centre of a circle. PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70°, find ∠TRQ.**

**Sol:**

We know that the radius and tangent are perpendicular at their point of contact ∠OTP = ∠OOP = 90°

Now, In quadrilateral OQPT

∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360° [Angle sum property of a quadrilateral] ∠QOT + 90° + 90° + 70° = 360°

250° + ∠QOT = 360°

∠QOT = 110°

We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.

∠TRQ = 12 and ∠QOT=55°

**Q.5: In the given figure, common tangents AB and CD to the two circles with centres O _{1} and O_{2 }intersect at E. Prove that AB = CD**

**Sol:**

We know that tangent segments to a circle from the same external point are congruent.

So, we have

EA = EC for the circle having centre O_{1}

And, ED = EB for the circle having centre O_{1}

Now, Adding ED on both sides in EA = EC, we get

EA + ED = EC + ED

=>EA + = EC + ED

=>AB = CD

**Q.6: If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70°, then find the measure of ∠POQ.**

**Sol:**

We know that the radius and tangent are perpendicular at their point of contact.

∠OPT = 90°

Now, ∠OPQ = ∠OPT – ∠TPQ = 90° – 70° = 20°

Since OP = OQ as both are radius

∠OPQ = ∠OQP = 20° (Angles opposite to equal sides are equal)

Now, In isosceles ⧍POQ

∠POQ + ∠OPQ + ∠OQP = 180° (Angle sum property of a triangle)

∠POQ = 180° – 20° – 20° = 140°

**Q.7: In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 4cm and 3cm respectively. If the area of triangle ABC = 21\(cm^{2}\) then find the lengths of sides AB and AC.**

**Sol:**

**Construction:** Join OA, 06, OC, OE perpendicular to AB at E and OF perpendicular to AC at F

We know that tangent segments to a circle from the same external point are congruent.

Now, we have

AE = AF, BD = BE = 4 cm and CD = CF = 3 cm

Now,

Area ⧍ABC=Area⧍B0C +Area ⧍AOB + Area of ⧍AOC

=>21 = 14 + 2x = 2x = 7

=> x = 3.5 cm

AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm

**Q.8: Two concentric circles are of radii 5cm and 3cm. Find the length of the chord of larger circle (in cm) which touches the smaller circle.**

**Sol:**

**Given:** Two circles have the same centre O and AB is a chord of the larger circle touching the smaller circle at C ; also. 0.4=5 cm and OC=3 cm.

In ⧍OAC, 0A^{2}=0C^{2}+AC^{2} + AC^{2 }= OA^{2}-OC^{2 }

=> AC^{2}=5^{2 }– 3^{2}

=>AC^{2}=25 – 9

=>AC^{2 }= 16

=>AC = 4 cm

AB = 2AC

AB = 2×4 = 8 cm

The length of the chord of the larger circle is 8 cm.

**Q.9: Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.**

Let AB be the tangent to the circle at point P with centre O.

**To prove:** PQ passes through the point O.

**Construction:** Join OP.

Through O, draw a straight line CD parallel to the tangent AB.

**Proof:** Suppose that PO doesn’t passes through point O.

PQ intersects CD at R and also intersects AB at P

AS, CD II AB, PO is the line of intersection,

∠ORP = ∠RPA (AJtemate interior angles)

But also, ∠RPA = 90° (OP perpendicular to AB)

∠ORP = 90°

∠ROP + ∠OPA = 180° (Co interior angles)

∠ROP + 90° = 180°

∠ROP = 90°

Thus, the ⧍ORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible. Hence, our supposition is wrong.

PQ passes through the point O.

**Q.10: In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120°, then prove that OR = PR + RQ.**

**Sol:**

**Construction:** Join PO and OQ

In ⧍POR and ⧍QOR

OP = OQ (Radii)

RP = RQ (Tangents from the external point are congruent)

OR = OR (Common)

By SSS congruency, ⧍POR = ⧍QOR

=> 2∠PRO = ∠QRO (C.P.C.T)

Now, ∠PRO + ∠QRO = ∠PRQ

=> 2∠PRO = 120°

∠PRO = 60°

Now, In ⧍PQR

cos60°= PROR

=>12 = PROR

=>OR = 2PR

=>OR = PR + PR

=>OR = PR + RQ

**Q.11: In the given figure, a circle inscribed in a triangle ABC touches the sides AB, BC and CA at points D, E and F respectively. If AB = 14cm cm, BC = 8cm and CA = 12cm. Find the length AD, BE and CF.**

**Sol:**

We know that tangent segments to a circle from the same external point are congruent.

Now, we have

AD = AF, BD = BE and CE = CF

Now, AD + BD = 14 cm………(1)

AF + FC = 12 cm

=> AD + FC = 12cm………(2)

BE + EC = 8 cm

=>BD + FC = 8cm ………..(3)

Adding all these we get

AD + BD + AD + FC + BD + FC = 34

=>2(AD + BD + FC) = 34

=>AD + BD + FC = 17 cm ……..(4)

Solving (1) and (4), we get

FC = 3 cm

Solving (2) and (4), we get

BD = 5 cm = BE

Solving (3) and (4), we get:

AD = 9 cm.

**Q.12: In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral.**

**Sol:**

We know that the radius and tangent are perperpendular at their point of contact

∠OBP = ∠OAP = 90°

Now, In quadrilateral AOBP

∠APB +∠AOB+ ∠OBP + ∠OAP = 360°

=> ∠APB + ∠AOB + 90° + 90° = 360°

∠APB + ∠AOB = 180°

Also, ∠OBP + ∠OAP = 180°

Since, the sum of the opposite angles of the quadrilateral is 180°

Hence, AOBP is a cyclic quadrilateral.

**Q.13:** **In two concentric circles, a chord of length 8cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5cm then find the radius of the smaller circle.**

We know that the radius and tangent are perpendicular at their point of contact since, the perpendicular drawn from the centre bisects the chord.

=> AP=PB=A62=4cm

In right triangle AOP

AO^{2} = OP^{2} + PA^{2}

=> 5^{2} = OP^{2 }+ 4^{2}

=> OP^{2} = 9

=> OP= 3cm

Hence, the radius of the smaller circle is 3 cm.

**Q.14: In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ.**

**Sol:**

We know that the radius and tangent are perpendicular at their point of contact.

∠OPT = 90°

Now, ∠OPO = ∠OPT – ∠QPT = 90° – 60° = 30°

Since OP = OQ as both are radius

∠OPQ= ∠OQP = 30° (Angles opposite to equal sides are equal)

Now, In isosceles ⧍POQ

∠POQ + ∠OPQ + ∠OQP = 180° (Angle sum property of a triangle)

=> ∠POQ = 180° – 30° – 30° = 120°

Now, ∠POQ + reflex ∠POQ = 360° (Complete angle)

=> reflex ∠POQ = 360° – 120° = 240°

We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.

∠PRQ = 12 reflex ∠POQ=120°

**Q.15: In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60° then find the measure of ∠OAB.**

**Sol:**

**Construction:** Join OB

We know that the radius and tangent are perpendicular at their point of contact

∠OBP = ∠OAP = 90°

Now, In quadrilateral AOBP

∠AOB + ∠OBP + ∠APB + ∠OAP = 360° [Angle sum property of a quadrilateral]

=> ∠AOB + 90° +60° +90° = 360°

=> 240° + ∠AOB = 360°

∠A0B = 120°

Now, In an isosceles triangle AOB

∠AOB + ∠OAB + ∠OBA = 180° [Angle sum property of a triangle]

=> 120° + 2∠OAB = 180°

=> ∠OAB = 30°