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**Q12. Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that âˆ APB = 2âˆ OAB**

**Sol.** We have PA and PB, the tangents to the circle and O is the centre of the circle.

âˆ´ PA = PB

â‡’âˆ 2 = âˆ 4 ...(1)

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

âˆ´âˆ OAP = 90Â°

â‡’ âˆ 1 + âˆ 2 = 90Â° ...(2)

â‡’ âˆ 2 = 90Â° âˆ’ âˆ 1

Now, in âˆ†ABP, we have:

âˆ´âˆ 2 + âˆ 3 + âˆ 4 = 180Â°

â‡’âˆ 2 + âˆ 3 + âˆ 2 = 180Â° [From (1)]

â‡’âˆ 2 + âˆ 3 = 180Â°

â‡’ 2 (90Â° âˆ’ âˆ 1) + âˆ 3 = 180Â° [From (2)]

â‡’ 180Â° âˆ’ 2 âˆ 1 + âˆ 3 = 180Â°

â‡’ 2 âˆ 1 = âˆ 3

â‡’ âˆ 3 = 2âˆ 1

â‡’âˆ APB = 2âˆ OAB

**Q13. ABC is an isosceles triangle, in which AB = AC, circumscribed about a circle. Show that BC is bisected at the point of contact.**

**Sol. **We know that the tangents to a circle from an external point are equal.

âˆ´ AD = AF

Similarly,

BD = BE

and CE = CF

Since AB = AC [Given]

â‡’ AB âˆ’ AD = AC âˆ’ AD

â‡’ AB âˆ’ AD = AC âˆ’ AF [âˆµ AD = AF]

â‡’ BD = CF ...(1)

But BF = BD and CF = CE

âˆ´ From (1), we have:

BE = CE

**Q14. If a, b, c are the sides of a right triangle where c is hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by **

**Sol. **Here, a, b and c are the sides of rt D ABC such that BC = a, CA = b and AB = c Let the circle touches the sides BC, CA, AB at D, E and F respectively.

= AE = AF and BD = BF

Also, CE = CD = r

âˆ´ AF = b â€“ r BF = a â€“ r

Now, AB = c â‡’ (AF + BF)

=(b â€“ r) + (a â€“ r)

â‡’ c = b + a â€“ 2r

â‡’ 2r = a + b â€“ c

**Q15. In a right Î” ABC, right angled at B, BC = 5 cm and AB = 12 cm. The circle is touching the sides of Î” ABC. Find the radius of the circle.**

**Sol.** Let the circle with centre O and radius â€˜râ€™ touches AB, BC and AC at P, Q, R, respectively.

Now,

AR = AP

âˆµ AP = AB â€“ BP = (12 â€“ r) cm

âˆ´ AR = (12â€“ r)cm

Similarly, CR = (5 â€“ r)cm

Now, using Pythagoras theorem in rt Î” ABC, we have

AC^{2} = AB^{2} + BC^{2}

â‡’ AC^{2} = 12^{2} + 5^{2}

â‡’ AC = 13 cm

But AC = AR + CR = (12 â€“ r) + (5 â€“ r)

â‡’ (12 â€“ r) + (5 â€“ r) = 13 cm

â‡’ 17 â€“ 2r = 13 cm

â‡’ 2 r = 17 â€“ 13 = 4 cm

â‡’ r = 4/2 = 2 cm

Thus, the radius of the circle is 2 cm.

**Q16. Prove that the parallelogram circumscribing a circle is a rhombus.**

**Sol. **Since ABCD is a â•‘ gm

âˆ´ AB = CD

and AD = BC

âˆµ Tangents from an external point to a circle are equal,

â‡’(AP + PB) + (RC + DR)

= (AS + DS) + (BQ + QC)

â‡’ AB + CD = AD + BC

â‡’ 2 AB =2 AD â‡’ AB = AD

â‡’ AB = AD = CD = BC

i.e., ABCD is a rhombus.

**Q17. In the following figure, OP is equal to diameter of the circle. Prove that ABP is an equilateral triangle.**

**Sol. **Since the tangent is perpendicular to the radius through the point of contact,

âˆ´ âˆ OAP = 90Â°

Let us join AB and AC.

In right âˆ†OAP, OP is the hypotenuse and C is the mid point of OP.

[âˆµ OP is a diameter of the circle (given)]

âˆ´ CA = CP = CO = Radius of the circle.

âˆ´ Î”OAC is an equilateral triangle.

Since all angles in an equilateral triangle are 60Â°,

âˆ´âˆ 1 = 60Â°

Now, in âˆ†OAP, we have

âˆ 1 + âˆ OAP + âˆ 2 = 180Â°

â‡’ 60Â° + 90Â° + âˆ 2 = 180Â°

â‡’âˆ 2 = 180Â° âˆ’ 90Â° âˆ’ 60Â° = 30Â°

Since PA and PB make equal angles with OP,

âˆ´âˆ 2= âˆ 3 â‡’âˆ 3 = 30Â°

âˆ´âˆ APB = âˆ 2 + âˆ 3

= 30Â° + 30Â° = 60Â°

Again, PA = PB.

â‡’ In Î”ABP,âˆ 4= âˆ 5

[Angles opposite to equal sides are equal]

Now, in Î”ABP, âˆ 4 + âˆ 5 + âˆ APB = 180Â°

â‡’âˆ 4 + âˆ 4 + âˆ APB= 180Â°

â‡’ 2âˆ 4 + âˆ 60Â° = 180Â°

â‡’ 2âˆ 4 = 180Â° âˆ’ 60Â° = 120Â°

Since, âˆ 4 = 60Â° âˆ 5 = 60Â°

âˆµ Î”ABP is an equilateral Î”.

âˆ APB = 60Â°

**Q18. Prove that the angle between the two tangents to a circle drawn from an external point is supplementary to the angle subtended by the line segment joining the points of contact at the centre.**

**Or**

**Two tangents PA and PB are drawn from an external point P to a circle with centre O. Prove that AOBP is a cyclic quadrilateral.**

**Sol. **

We have tangents PA and PB to the circle from the external point P. Since a tangent to a circle is perpendicular to the radius through the point of contact,

âˆ´ âˆ 2 = 90Â° and âˆ 4 = 90Â°

Now, in quadrilateral OAPB,

âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 = 360Â°

â‡’âˆ 1 + 90Â° + âˆ 3 + 90Â° = 360Â°

â‡’âˆ 1 + âˆ 3 = 360Â° âˆ’ 90Â° âˆ’ 90Â°

= 180Â°

i.e., âˆ 1 and âˆ 3 are supplementary angles.

â‡’âˆ AOB and âˆ APB are supplementary

â‡’ AOBP is a cyclic quadrilateral.

Q19. Two equal circles, with centres O and Oâ€², touch each other at X. OOâ€² produced meets the circle with centre Oâ€² at A. AC is tangent to the circle with centre O, at the point C. Oâ€²D is perpendicular to AC. Find

**Sol.** AC is tangent to circle with centre O at C (given)

âˆ ACO = 90Â° â‡’ âˆ† ACO is a rt Î”

âˆ ADOâ€² = 90Â° [ Ã¤ Oâ€²D âŠ¥ AC] â‡’ Î”

ADOâ€² is a rt Î”

But, AOâ€² = r, Oâ€²X = r and OX = r â‡’ AO = 3r

From (1) and (2), we get

**Q20. Out of two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.**

Sol. Let the given chord AC of the larger circle touch the smaller circle at L. âˆµ AC is a tangent at L to the smaller circle with centre O

âˆ´ OL âŠ¥ AC

Also AC is a chord of the bigger circle

âˆ´

Now, in rt. Î”OAL,

OL^{2} = OA^{2} â€“ AL^{2}

or OL^{2} = 5^{2} â€“ 4^{2}

= (5 + 4) (5 â€“ 4)

= 9 Ã— 1 = 9

â‡’

Thus, the radius of the inner circle is 3 cm.

**Q21. In the figure, O is the centre of a circle of radius 5cm. T is a point such that OT = 13cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle. **

Sol. O is centre of the circle and PT is a tangent to circle

âˆ´ âˆ OPT = 90Â° â‡’ âˆ† OPT is a rt âˆ† using Pythagoras theorem

OT^{2} = OP^{2} + PT^{2} or 132 = 52 + PT2

â‡’ PT2 =132 â€“ 52

â‡’ PT =

Let AP = AE = x

[Tangent to a circle from an external point are equal]

â‡’ AT = PT â€“ AP = (12 â€“ x) cm

[âˆµ AB is a tangent to the circle at E and OE is a radius]

âˆ´ âˆ OEA = 90Â° â‡’ âˆ AET = 90Â°

Î” AET is a rt âˆ†

â‡’ AT^{2} = AE^{2} + ET^{2}

or (12 âˆ’ x)^{2} = x^{2} + (13 âˆ’ 5)^{2}

â‡’ 144âˆ’ 24x + x^{2} = x^{2} + 64 or 24x = 80

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