question_answer

If OA and OB are the tangents form 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 OACD 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 :

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