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**SHORT ANSWER TYPE QUESTIONS**

**Q1. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.**

**Sol.**

Let NM be a chord of a circle with centre C.

Let the tangents at M and N meet at O.

âˆµ OM is a tangent at M

âˆµ âˆ OMC = 90Â° ...(1)

Similarly âˆ ONC = 90Â° ...(2)

Since, CM = CN [Radii of the same circle]

âˆµ In D CMN, âˆ 1= âˆ 2

From (1) and (2), we have

âˆ OMC â€“âˆ 1 = âˆ ONC â€“âˆ 2

â‡’ âˆ OML = âˆ ONL

Thus, tangents make equal angles with the chord.

**Q2. Two concentric circles have a common centre O. The chord AB to the bigger circle touches the smaller circle at P. If OP = 3 cm and AB = 8 cm then find the radius of the bigger circle.**

**Sol. âˆµ **AB touches the smaller circle at P.

âˆ´ OP âŠ¥ AB â‡’ âˆ OPA = 90Â°

Now, AB is a chord of the bigger circle.

Since, the perpendicular from the centre to a chord, bisects the chord,

âˆ´ P is the mid-point of AB

â‡’

In right âˆ† APO, we have

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

â‡’ AO^{2} = 3^{2} + 4^{2}

â‡’ AO^{2} = 9 + 16 = 25 = 5^{2}

â‡’

Thus, the radius of the bigger circle is 5 cm.

**Q3. In the given figure, O is the centre of the circle and PQ is a tangent to it. If its circumference is 12Ï€ cm, then find the length of the tangent.**

**Sol. **âˆµ Circumference of the circle = 12Ï€ cm

âˆ´ 2Ï€ r =12Ï€

[âˆµ r is the radius of the circle]

â‡’

â‡’ Radius of the circle = 6 cm = OQ

Since a tangent to circle is perpendicular to the radius through the point of contact,

âˆ´ âˆ OQP = 90Â°

Now, in rt Î” OQP, we have:

OQ^{2} + QP^{2} = OP^{2}

â‡’ 6^{2} + QP^{2} =10^{2}

â‡’ QP^{2} = 10^{2} âˆ’ 6^{2} = (10 âˆ’ 6) (10 + 6) = 4 Ã— 16 = 64 = 8^{2}

â‡’

Thus, the length of the tangent is 8 cm.

**Q4. Given two concentric circles of radii 10 cm and 6 cm. Find the length of the chord of the larger circle which touches the other circle.**

**Sol. **The chord AB touches the inner circle at P.

âˆ´ AB is tangent to the inner circle.

â‡’ OP âŠ¥ AB

[âˆµ O is the centre and OP is radius through the point of contact P]

âˆ´ âˆ OPB = 90Â°.

Now, in right âˆ† OPB, we have:

OP^{2} + PB^{2} = OB^{2}

â‡’ 6^{2} + PB^{2} = 10^{2}

â‡’ PB^{2} = 10^{2} âˆ’ 6^{2}

= (10 âˆ’ 6) Ã— (10 + 6)

â‡’ PB^{2} = 4 Ã— 16

â‡’ PB^{2} = 64 = 8^{2}

â‡’

âˆµ The radius perpendicular to a chord bisects the chord.

âˆ´ P is the mid-point of AB

âˆ´ AB = 2 Ã— PB = 2 Ã— 8 = 16 cm.

**Q5. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that âˆ PTQ = 2 âˆ OPQ.**

**Sol.** âˆµ Tangent to a circle is perpendicular to the radius through the point of contact.

In quadrilateral. OPTQ,

âˆ OPT + âˆ OQT + âˆ POQ + âˆ PTQ = 360Â°

or 90Â° + 90Â° + âˆ POQ + âˆ PTQ = 360Â°

â‡’ âˆ POQ + âˆ PTQ = 360Â° âˆ’ 90Â° âˆ’ 90Â° = 180Â° ...(1)

In Î” OPQ, âˆ 1 + âˆ 2 + âˆ POQ = 180Â° ...(2)

Since OP =OQ [Radii of the same circle]

â‡’ âˆ 1 = âˆ 2 [Angles opposite to equal sides]

âˆ´ âˆ OPT = 90Â° = âˆ OQT

âˆ´ From (2), we have

âˆ 1 + âˆ 1 + âˆ POQ = 180Â°

â‡’ 2 âˆ 1 + âˆ POQ = 180Â° ...(3)

From (1) and (3), we have

2 âˆ 1 + âˆ POQ = âˆ POQ + âˆ PTQ

â‡’ 2 âˆ 1= âˆ PTQ

â‡’ 2 âˆ OPQ = âˆ PTQ.

**Q6. In the figure, the incircle of âˆ† ABC touches the sides BC, CA and AB at D, E and F respectively. If AB = AC, prove that BD = CD.**

**Sol.** Since the lengths of tangents drawn from an external point to a circle are equal,

âˆ´ We have

AF = AE

BF = BD

CD = CE

Adding them, we get

(AF + BF) + CD = (AE + CE) + BD

â‡’ AB + CD = AC + BD

But AB = AC (Given)

âˆ´ CD = BD.

**Q7. A circle is touching the side BC of a D ABC at P and touching AB and AC produced at Q and R. Prove that: **

**Sol.** Since, the two tangents drawn to a circle from an external point are equal.

âˆµ AQ = AR ...(1)

Similarly, BQ = BP ...(2)

and CR = CP ...(3)

Now, Perimeter of Î” ABC

= AB + BC + AC

= AB + (BP + PC) + AC

= AB + (BQ + CR) + AC

[From (2) and (3)]

= (AB + BQ) + (CR + AC)

= AQ + AR

= AQ + AQ [From (1)] = 2AQ

â‡’

**Q8. In the given figure, the radii of two concentric circles are 13 cm and 8 cm. AB is diameter of the bigger circle. BD is the tangent to the smaller circle touching it at D. Find the length AD. **

**Sol.** Join AE and OD

âˆµ âˆ ODB = 90Â°

[BE is a tangent at D and OD is a radius] and âˆ AEB = 90Â° [AB is diameter so âˆ AEB is an angle in semicircle so âˆ AEB = 90Â°]

âˆ´ OD â•‘ Î‘Î• and âˆ†BEA are similar by AA-Similarity

So

Again in rt Î” AED,

**Q9. In two concentric circles, a chord of the larger circle touches the smaller circle. If the length of this chord is 8 cm and the diameter of the smaller circle is 6 cm, then find the diameter of the larger circle.**

**Sol.** Let the common centre be O. Let AB be the chord of the larger circle.

âˆ´ AB = 8 cm

And CD is the diameter of the smaller circle i.e., CD = 6 cm

â‡’

Join OA. D is the point of contact.

âˆ´ OD âŠ¥ AB

â‡’ D is the mid point of AB

â‡’ AD = 4 cm Now, in right Î”ADO, we have:

AO^{2} = AD^{2} + OD^{2}

= 42 + 3^{2} = 16 + 9 = 25 = 5^{2}

â‡’ AO = 5 cm

â‡’ 2AO = 2(5 cm) = 10 cm

âˆ´ The diameter of the bigger circle is 10 cm.

**Q10. In the following figure, PA and PB are two tangents drawn to a circle with centre O, from an external point P such that PA = 5 cm and âˆ APB = 60Â°. Find the length of chord AB.**

**Sol.** Since the tangents to a circle from an external point are equal,

âˆ´ PA = PB = 5 cm

In Î”PAB, we have

âˆ PAB = âˆ PBA [âˆµ PA = PB]

âˆ´âˆ PAB + âˆ PBA + âˆ APB = 180Â°

â‡’âˆ PAB + âˆ PAB + 60Â° = 180Â°

â‡’ 2 âˆ PAB + 60Â° = 180Â°

â‡’ 2 âˆ PAB = 180Â°âˆ’ 60Â°

= 120Â°

â‡’ âˆ PAB = 60Â°

â‡’ Each angle of âˆ†PAB is 60Â°.

â‡’ Î”PAB is an equilateral triangle. âˆ´

PA = PB

= AB = 5 cm

Thus, AB = 5 cm

**Q11. In the following figure, AB is a chord of length 9.6 cm of a circle with centre O and radius 6 cm.**

The tangents at A and B intersect at P. Find the length PA.

**Sol.**

Join OB.

Let PA = x cm and PR = y cm Since, OP is perpendicular bisector of AB

âˆ´ AR = BR = 9.6/2 = 4.8 cm

Now, in rt âˆ†OAR, we have:

OA^{2} = OR^{2} + AR^{2}

[By Pythagoras theorem]

â‡’ OR^{2} = OA^{2} âˆ’ AR^{2}

= 62 âˆ’ (4.8)^{2} = (6 âˆ’ 4.8) Ã— (6 + 4.8) = 1.2 Ã— 10.8

â‡’ = 12.96

OR = 3.6 cm.

Again, in right Î”OAP,

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

OP^{2} = (AR^{2} + PR^{2}) + OA^{2}

[âˆµ AP^{2} = AR^{2} + PR^{2}]

â‡’ (y + 3.6)2 = (4.8)^{2} + y^{2} + 6^{2}

â‡’y^{2} + 12.96 + 7.2 y = 23.04 + y^{2} + 36

â‡’ 7.2 y = 46.08

â‡’

â‡’ PR = 6.4 cm

Now, AP^{2} = AP^{2} + PR^{2}

= (4.8)^{2} + (6.4)^{2} = 23.04 + 40.96 = 64

â‡’

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