In Q.1 to 3, choose the correct option and give justification.
1.   From a point Q, the length of the tangent to a circle is 4 cm and the distance of Q from the centre is 5 cm.
	The radius of the circle is
	(A)   7cm                                  (B)   3cm
	(C)   5cm                                (D)   4.5cm
	
Answer


Given : QP= 4 cm
        QO=  5 cm
		Since OPQ is a right angle triangle
		Hence, $OQ^2= OP^2+PQ^2$
				$OP = √(OQ^2-PQ^2)$
				$OP = √(5^2-4^2)$
				$OP = √{(5-4)(5+4)}$
				$OP = √{9}$
				$OP = 3$ $cm$

	
2.   In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 90°, then ∠PTQ is equal to
	(A)   60°                                    (B)   70° 
	(C)   80°                                    (D)  90°


Answer
			Since TP and TQ are tangent to the circle, therefore  ∠OPT = 90° and  ∠OQT = 90°
			Also  ∠POQ + ∠PTQ = 180°
			      90° + ∠PTQ = 180°
			     ∠PTQ = 180° - 90°
			     ∠PTQ =  90°

3.   If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 60°, then ∠POA is equal to
(A)   50°                   (B)   60°                               (C)   70°                    (D)   80°                            

Solution:

In trianlge AOP, 
∠APO + ∠POA +∠OAP = 180°
30° + ∠POA +90° = 180°
∠POA +120° = 180°
∠POA = 60°


4.   Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Proof

5.   Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

6.   The length of a tangent from a point A at distance 15 cm from the centre of the circle is 12 cm. 
	Find the radius of the circle.
Solution	
	

Given : AB= 12 cm
        OA=  15 cm
		Since OAB is a right angle triangle
		Hence, $OA^2= AB^2+OB^2$
				$OB = √(OA^2-AB^2)$
				$OB = √(15^2-12^2)$
				$OB = √{(15-12)(15+12)}$
				$OB = √{(3)(27)}$
				$OB = 9$ $cm$
				

7.   Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which 
	touches the smaller circle.
Solution	
	 

Radius of larger circle= OA = 5 cm.
Radius of smaller circle=OP= 3 cm.

Since the cord AB of larger circle C1  is tangent to circle C2, so OP is perpendicular to AB .

SO in triangle APO, $OA^2 =AP^2+OP^2$
            $AP=√(OA^2 -OP^2)$
            $AP=√(5^2 -3^2)$
            $AP=√{(5+3)(5 -3)}$
            $AP=√{(8)(2)}$
            $AP=√{16}$
            $AP=4$ $cm$
Since OP is the line bisector of line AB, so AP=PB

So the length of cord = AB= $AP+PB$
						  = $2  ×AP$
						  = $2  ×4$
						  = $8$ $cm$

8.   A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. ). Prove that AB + CD =AD + BC.

	 

Solution:

Since AS and AP are tangent to the circle from point A, 
			so  AS=AP  ..............(1)

Similiarly, BP=BQ, .................(2)
			CR=CQ, ..................(3)
			DR=DS ....................(4)
	
	So, AB+CD => (AP+BP) + (CR+DR)
	          => (AS+BQ) +(CQ+DS)
			  =>(AS+DS)+(BQ+CQ)
			  =>AD+ BC = R.H.S 
			  

9.   In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C
	intersecting XY at A and X′Y′at B. Prove that ∠AOB = 90°.
	

	
10.   Prove that the angle between the two tangents drawn from an external point to a circle is  supplementary  to  the  angle 
	 subtended  by  the  line-segment  joining  the  points  of contact at the centre.
	 
11.   Prove that the parallelogram circumscribing a circle is a rhombus.

12.   Atriangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by
	the point of contact  D  are  of  lengths  8  cm  and  6  cm respectively (see Fig. 10.14). Find the sidesAB and AC.
	
13.   Prove  that  opposite  sides  of  a  quadrilateral circumscribing a circle subtend supplementary angles at the centre
	of the circle.