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**Short Answer Type Questions**

**Q.1. In fig., AB is a chord of length 8 cm of a circle of radius 5 cm. The tangents to the circle at A and B intersect at P. Find the length of AP. [Delhi 2019, CBSE 2018 (C)]****Ans. **Given: PA and PB are tangents to a circle.

Chord AB = 8 cm and radius OA = 5 cm

Construction: Join OP intersecting AB at M.

To find: Length AP

Solution: OP is perpendicular bisector of AB intersecting AB at M.

In ΔOMA

OA^{2} = OM^{2 }+AM^{2}

(5)^{2} = OM^{2} + (4)^{2}

25 = OM^{2} + 16

OM^{2} = 9 ⇒ OM - 3 cm

In ΔOAP, ∠OAP = 90°

(radius is perpendicular to tangent)

OP^{2} = OA^{2} + AP^{2}

AP^{2} = OP^{2} - OA^{2} ...(i)

In ΔAMP, ∠AMP = 90°

AP^{2} = AM^{2} + PM^{2}

= AM^{2} + (OP - OM)^{2} ...(ii)

Equating (i) and (ii), we get

OP^{2} - OA^{2} = AM^{2} + (OP - OM)^{2}

⇒ OP^{2} - (5)^{2} = (4)^{2} + (OP - 3)^{2}

⇒OP^{2} - 25 = 16 + OP^{2} + 9 - 6(OP)

⇒ - 6(OP) = - 50

∴ Eq (i) becomes**Q.2. Prove that the lengths of tangents drawn from an external point to a circle are equal. [Delhi 2018] ****Ans. **Given: In circle, O is the centre. P is an external point and PA and PB are the tangents drawn.

To prove: PA = PB

Construction: Join OA, OB and OP.

Proof: Since, PA and PB are the tangents and OA and OB are the radii of a circle.

∴ OA ⊥ PA and OB ⊥ PB

[Tangent to a circle makes angle 90° with the radius at die point of contact]

⇒ ∠OAP = ∠OBP = 90°

Now, in ΔOAP and ΔOBP,

OA = OB (Radii)

OP = OP (Common)

∠OAP = ∠OBP (Each 90°)

ΔOAP ≌ ΔOBP

[By RHS congruence rule]

PA = PB [By CPCT]

Hence, proved.**Q.3. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. [Delhi 2017]****Ans. **Given: A circle with centre O.

An external point T from which TP and TQ are two tangents to the circle.

To prove: **∠**PTQ = 2**∠**OPQ

Proof: Let**∠**PTQ = θ

Now TP = TQ [Lengths of tangent segments from an external point to a circle are equal]

∴ ΔTPQ is an isosceles triangle.

⇒ **∠**TPQ = **∠**TQP [Angles opposite to equal sides of a triangle are equal]

In triangle TPQ**∠**PTQ + **∠**TPQ + **∠**TQP = 180°

[Angle sum property of triangles]

⇒ θ + 2**∠**TPQ = 180°

Also, **∠**OPT = 90° [Angle between tangent and radius through the point of contact]

So, **∠**OPQ - **∠**ZOPT - **∠**TPQ**Q.4. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60°, then find the length of OP. [Delhi 2017]****Ans. **Given: Radius of circle = a

∠QPR = 60°

In ΔPOQ and ΔPOR,

OQ = OR (radii), ∠Q = ∠R - (Each 90°), PQ = PR

(∵ PR and PQ are tangents from an external point P)

Now, in right triangle OQP,

OQ/OP = sin 30º**Q.5. In the given fig., 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 length of PS [Delhi 2017 (C)]****Ans. **PR and PQ are tangents to the external point P.

∴ PR = PQ

∴ PQ = 5 c m (∵ PR = 5 cm)

PS = 2PQ (∵ OQ ⊥ PS)

= 2 x 5

= 10 cm**Q.6. In the given Fig., O is the centre of the circle, PQ is a chord and PT is tangent to the circle at P. If ∠POQ = 70°, find ∠TPQ. [AI 2017 (C)]****Ans. **Given: O is centre of the circle, PQ is chord and PT is tangent at P.

To find: ∠TPQ

Solution: In ΔOPQ

∠POQ + ∠OPQ + ∠OQP - 180°

⇒ 70° + ∠OPQ + ∠OPQ = 180°

(∵ OP = OQ, radii of the circle)

⇒ 70° + 2∠OPQ = 180°

⇒ 2∠OPQ = 110°

⇒ ∠OPQ = 55°

OP is perpendicular to the tangent at P.

∠OPT = 90°

⇒ ∠OPQ + ∠TPQ - 90°

⇒ 55° + ∠TPQ =90°

⇒ ∠TPQ = 90° - 55°

∠TPQ = 35°**Q.7. In the figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD. [Delhi 2017]****Ans. **Construction: Join AD and BC**Proof:** We know that the lengths of tangents from an external point to a circle are equal.

So, if A is an external point for circle having centre O', then

AB = AD ...(i)

If ‘C’ is an external point for circle with centre O', then

CB = CD ...(ii)

Also, B is an external point for circle with centre O,

∴ BC = AB ...(iii)

D is an external point for circle with centre O.

So DA = DC ...(iv)

So, from (i), (ii), (iii) and (iv), we get

AB = BC = CD

⇒ AB = CD Hence proved**Q.8. ****Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord. [AI 2017] Sol.** Let AB be the chord of the circle with centre O.

Let PQ and RS respectively be the tangents at the end points A and B of the chord.

Join OA and OB.

ΔOAB is an isosceles as OA = OB = r

[radii of the same circle]

∴ ∠OAB = ∠OBA ...(i)

[Angles opposite to equal sides of a triangle are equal]

Now ∠OAP = ∠OBR ... (ii)

[90° each angle between tangents and radius at the point of contact]

Adding equation (i) and equation (ii), we get

∠OAB + ∠OAP = ∠OBA + ∠OBR

⇒ ∠BAP = ∠ABR

Also, ∠OAQ = ∠OBS [90° each] ...(iii)

Subtracting equation (i) from equation (iii), wc got

∠OAQ - ∠OAB = ∠OBS - ∠OBA

∠BAQ = ∠ABS

Hence proved.

OQ is perpendicular to the tangent PQ. (radius is perpendicular to the tangent at the point of contact).

In ΔOPQ

∠POR = ∠1 + ∠RQP

(exterior angle is equal to the .......... sum of interior opposite angles)

130° = ∠1 + 90°

∠1 = 130° - 90° = 40° ...(i)

∠ROT is the angle subtended by arc RT at centre and ∠RST is the angle subtended by arc RT at point S on circumference.

∴

Adding (i) and (ii), we get

∠1 + ∠2 = 40° + 65° = 105°

To find: ∠PKL

Solution In ΔOKL, OL = OK (radii of a circle)

∴ ∠OLK = ∠OKL = 30°

(angle opposite to equal sides are equal)

OK is perpendicular to the tangent PQ at K.

∴ ∠OKP = 90°

⇒ ∠OKL + ∠LKP = 90°

⇒ 30° + ∠PKL = 90°

∴ ∠PKL = 60°

84 = 2(AB + AC + BC)

42 = AB + AC + BC

⇒ AC + BC = 28 cm ...(i) [∵ AB = 14 cm]

Now, AP = AR ; BP = BQ

and CQ = CR = x (say)

(Length tangents from common exterior point)

So, AR = 6 cm, BQ = 8 cm, CQ = x

From eq (z) we have

AR + CR + CQ + BQ = 28

6 + x + x + 8 = 28

2x + 14 = 28

2x = 14

⇒ x = 7

∴ AC = AR + RC

= 6 + 7 = 13 cm

BC = BQ + QC

= 8 + 7 = 15 cm

⇒ 12 = PC + 3 ⇒ PC = 9 cm

∵ PA =PB ⇒ PA - AC = PB - BD ⇒ PC = PD

∴ PD = 9 cm

Hence, PC + PD = 18 cm**Q.13. The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC. [CBSE (AI) 20 14]****OR****In Fig.8.32, if AB =AC, prove that BE = EC. [CBSE (E) 2017]****[Note: D, E, F replace by F, D, E]****Ans. **Given, AB = AC (In Fig 8.37)

We have, BE + AF = AE + CE ......(i)

AB, BC and CA are tangents to the circle at. F, D and E respectively.

∴ BF = BD, AE = AF and CE = CD ....(ii)

From (i) and (ii)

BD + AE = AE + CD (∵ AF = AE)

⇒ BD = CD**Q.14. ****A quadrilateral ABCD is drawn to circumscribe a circle (Fig. 8.55). Prove that AB + CD = AD + BC. [NCERT, CBSE (AI) 2016]****OR****A circle touches all the four sides o f a quadrilateral ABCD. Prove that AB + CD = BC + DA [CBSE (AI) 2017, Delhi 2017 (C)]****Ans. **Since lengths of two tangents drawn from an external point of circle are equal,

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

(where P, Q, R and S are the points of contact)

Adding all these, we have

(AP + BP) + (CR + RD) = (BQ + CQ) + (DS + AS)

⇒ AB + CD = BC + DA

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