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Q.

From a point ‘P’ on the line $2 x + y + 4 = 0$ ; which is nearest to the circle $x^{2} + y^{2} - 12 y + 35 = 0$ tangents are drawn to given circle. The area of quadrilateral $P A C B$ (where ‘C’ is the center of circle and $P A & P B$ are the tangents.) is :

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a

8

b

$\sqrt{19}$

c

$\sqrt{110}$

d

9

answer is C.

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Detailed Solution

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$T h e p e r p e n d i c u l a r l i n e o f 2 x + y + 4 = 0 a n d p a s \sin g t h r o u g h (0, 6) i s x - 2 y = - 12 s o l v e t h e a b o v e t w o e q u a t i o n s w e w i l l g e t (- 4, 4) T h e l e n g t h o f t h e \tan g e n t f r o m (- 4, 4) t o t h e c i r c l e = \sqrt{19} A r e a o f P A B C = 2 (\frac{1}{2} . 1 . \sqrt{19})$

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