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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 Î”OMA

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

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

25 = OM^{2} + 16

OM^{2} = 9 â‡’ OM - 3 cm

In Î”OAP, âˆ OAP = 90Â°

(radius is perpendicular to tangent)

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

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

In Î”AMP, âˆ AMP = 90Â°

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}

â‡’ OP^{2} - (5)^{2} = (4)^{2} + (OP - 3)^{2}

â‡’OP^{2} - 25 = 16 + OP^{2} + 9 - 6(OP)

â‡’ - 6(OP) = - 50

âˆ´ 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.

âˆ´ OA âŠ¥ PA and OB âŠ¥ PB

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

â‡’ âˆ OAP = âˆ OBP = 90Â°

Now, in Î”OAP and Î”OBP,

OA = OB (Radii)

OP = OP (Common)

âˆ OAP = âˆ OBP (Each 90Â°)

Î”OAP â‰Œ Î”OBP

[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 âˆ PTQ = 2âˆ 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: **âˆ **PTQ = 2**âˆ **OPQ

Proof: Let**âˆ **PTQ = Î¸

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

âˆ´ Î”TPQ is an isosceles triangle.

â‡’ **âˆ **TPQ = **âˆ **TQP [Angles opposite to equal sides of a triangle are equal]

In triangle TPQ**âˆ **PTQ + **âˆ **TPQ + **âˆ **TQP = 180Â°

[Angle sum property of triangles]

â‡’ Î¸ + 2**âˆ **TPQ = 180Â°

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

So, **âˆ **OPQ - **âˆ **ZOPT - **âˆ **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 60Â°, then find the length of OP. [Delhi 2017]****Ans. **Given: Radius of circle = a

âˆ QPR = 60Â°

In Î”POQ and Î”POR,

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

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

Now, in right triangle OQP,

OQ/OP = sin 30Âº**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.

âˆ´ PR = PQ

âˆ´ PQ = 5 c m (âˆµ PR = 5 cm)

PS = 2PQ (âˆµ OQ âŠ¥ 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 âˆ POQ = 70Â°, find âˆ TPQ. [AI 2017 (C)]****Ans. **Given: O is centre of the circle, PQ is chord and PT is tangent at P.

To find: âˆ TPQ

Solution: In Î”OPQ

âˆ POQ + âˆ OPQ + âˆ OQP - 180Â°

â‡’ 70Â° + âˆ OPQ + âˆ OPQ = 180Â°

(âˆµ OP = OQ, radii of the circle)

â‡’ 70Â° + 2âˆ OPQ = 180Â°

â‡’ 2âˆ OPQ = 110Â°

â‡’ âˆ OPQ = 55Â°

OP is perpendicular to the tangent at P.

âˆ OPT = 90Â°

â‡’ âˆ OPQ + âˆ TPQ - 90Â°

â‡’ 55Â° + âˆ TPQ =90Â°

â‡’ âˆ TPQ = 90Â° - 55Â°

âˆ TPQ = 35Â°**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 â€˜Câ€™ is an external point for circle with centre O', then

CB = CD ...(ii)

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

âˆ´ 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

â‡’ 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.

Î”OAB is an isosceles as OA = OB = r

[radii of the same circle]

âˆ´ âˆ OAB = âˆ OBA ...(i)

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

Now âˆ OAP = âˆ OBR ... (ii)

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

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

âˆ OAB + âˆ OAP = âˆ OBA + âˆ OBR

â‡’ âˆ BAP = âˆ ABR

Also, âˆ OAQ = âˆ OBS [90Â° each] ...(iii)

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

âˆ OAQ - âˆ OAB = âˆ OBS - âˆ OBA

âˆ BAQ = âˆ ABS

Hence proved.

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

In Î”OPQ

âˆ POR = âˆ 1 + âˆ RQP

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

130Â° = âˆ 1 + 90Â°

âˆ 1 = 130Â° - 90Â° = 40Â° ...(i)

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

âˆ´

Adding (i) and (ii), we get

âˆ 1 + âˆ 2 = 40Â° + 65Â° = 105Â°

To find: âˆ PKL

Solution In Î”OKL, OL = OK (radii of a circle)

âˆ´ âˆ OLK = âˆ OKL = 30Â°

(angle opposite to equal sides are equal)

OK is perpendicular to the tangent PQ at K.

âˆ´ âˆ OKP = 90Â°

â‡’ âˆ OKL + âˆ LKP = 90Â°

â‡’ 30Â° + âˆ PKL = 90Â°

âˆ´ âˆ PKL = 60Â°

84 = 2(AB + AC + BC)

42 = AB + AC + BC

â‡’ AC + BC = 28 cm ...(i) [âˆµ 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

â‡’ x = 7

âˆ´ AC = AR + RC

= 6 + 7 = 13 cm

BC = BQ + QC

= 8 + 7 = 15 cm

â‡’ 12 = PC + 3 â‡’ PC = 9 cm

âˆµ PA =PB â‡’ PA - AC = PB - BD â‡’ PC = PD

âˆ´ 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.

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

From (i) and (ii)

BD + AE = AE + CD (âˆµ AF = AE)

â‡’ 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)

â‡’ AB + CD = BC + DA

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