question_answer

If OA and OB are the tangents from the origin to the circle \[{{x}^{2}}+{{y}^{2}}+2\,gx+2\,fy+c=0\] 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 :

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