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If \mathrm{O A}  and \mathrm{O B} are the tangents from the origin to the circle \mathrm{x^2+y^2+2 g x+2 f y+c=0}, and C is the centre of the circle, the area of the quadrilateral \mathrm{O A C B} is
 

Option: 1

\frac{1}{2} \sqrt{\mathrm{c}\left(\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}\right)}


Option: 2

\sqrt{\mathrm{c}\left(\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}\right)}


Option: 3

\mathrm{ c \sqrt{\left(g^2+f^2-c\right)} }


Option: 4

\frac{\sqrt{\left(\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}\right)}}{\mathrm{c}}


Answers (1)

best_answer

Since \mathrm{O A=O B} and \mathrm{C A=C B}, the diagonal \mathrm{O C} divides the quadrilateral \mathrm{O A C B} in two equal right angled triangles, \mathrm{O A C} and \mathrm{O B C}.

Therefore, the area of the quadrilateral \mathrm{O A C B} is\mathrm{2 (area ~ of ~ triangle \mathrm{~OA}\mathrm{C})=2 \times\left (\frac{1 }{2} \right ) \mathrm{OA} \times \mathrm{AC} }
\mathrm{=\sqrt{0+0+2 \mathrm{~g} \times 0+2 \mathrm{f} \times 0+\mathrm{c}} \sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{\mathrm{c}\left(\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}\right)} }

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rishi.raj

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