A is the centre of the circle x 2 + y 2 − 2 x − 4 y − 20 = 0 . If the tangents drawn at the points B ( 1 , 7 ) and D ( 4 , − 2 ) on the circle meet at the point C , then area of the quadrilateral A BC D (in square units) is

Question

A is the centre of the circle x 2 + y 2 − 2 x − 4 y − 20 = 0 . If the tangents drawn at the points B ( 1 , 7 ) and D ( 4 , − 2 ) on the circle meet at the point C , then area of the quadrilateral A BC D (in square units) is, 75, 64, 56, 45

MARKS App · Accepted Answer

Equation of the given circle is Now, equation of tangent at the point B ( 1 , 7 ) on the circle is $ x + 7 y − ( x + 1 ) − 2 ( y + 7 ) − 20 = 0 ⇒ 5 y = 35 im g src = " h ttp s : // c d n − q u es t i o n − p oo l . g e t ma r k s . a pp / p y q / a p e ​ am ce t / tN − U P b Qp I Z r h U o u z 70 oy 5 iiV S 50 O x 4 r L H 8 u i L 4 g V 5 oo . or i g ina l . f u ll s i ze . p n g " > b r > S imi l a r l y , e q u a t i o n o f t an g e n t a tt h e p o in t D(4,-2) o n t h ec i rc l e i s 4 x − 2 y − ( x + 4 ) − 2 ( y − 2 ) − 20 = 0 im g src = " h ttp s : // c d n − q u es t i o n − p oo l . g e t ma r k s . a pp / p y q / a p e ​ am ce t / m 0 G 6 Ya e O W H U NN 7 TMGX U o R 9 k A 7 h 4 T h w A F g n 4 S b A S m Z uU . or i g ina l . f u ll s i ze . p n g " > b r > N o w , p o in t o f in t ersec t i o n o f t an g e n t s ( ii ) an d ( iii ) i s C(16,7) . N o w , a re a o f re q u i re d q u a d r i l a t er a l A B C D im g src = " h ttp s : // c d n − q u es t i o n − p oo l . g e t ma r k s . a pp / p y q / a p e ​ am ce t / G i 62 f − 3 LU 2 N 85 N o b 3 Nqq K d U r RG vvu X L z MM 6 TT 295 − yo . or i g ina l . f u ll s i ze . p n g " > b r > ​ = 2 × Area of △ A BC = 2 × 2 1 ​ r S 1 ​ ​ [where r = radius of circle ( i ) = 1 + 4 + 20 ​ = 5 and S 1 ​ ​ = 256 + 49 − 32 − 28 − 20 ​ = 225 ​ = 15 ] = 5 × 15 = 75 ​ $ Hence, option (a) is correct.

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