question_answer

. If OA and OB are the tangents from the origin to the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]and C is the centre of the circle, the area of the quadrilateral \[OACB\]is

A)            \[\frac{1}{2}\sqrt{c({{g}^{2}}+{{f}^{2}}-c)}\]                

B)            \[\sqrt{c({{g}^{2}}+{{f}^{2}}-c)}\]

C)            \[c\sqrt{{{g}^{2}}+{{f}^{2}}-c}\]                                      

D)            \[\frac{\sqrt{{{g}^{2}}+{{f}^{2}}-c}}{c}\]

Correct Answer: B

Solution :

           Area of quadrilateral\[=2\]   [area of \[\Delta OAC\]]                    \[=2.\frac{1}{2}OA\ .\ AC=\sqrt{{{S}_{1}}}.\,\sqrt{{{g}^{2}}+{{f}^{2}}-c}\]            Point is (0, 0)\[\Rightarrow {{S}_{1}}=c\]. \[\therefore \] Area\[=\sqrt{c}.\sqrt{{{g}^{2}}+{{f}^{2}}-c}\].