Question

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. $\frac{\sqrt{g^2+f^2-c}}{c}$

Answer 1

Answer: (B)

$\sqrt{c(g^2+f^2-c)}$

Since $OA=OB$ and $CA=CB$

The diagonal $OC$ divides the quadrilateral $OACB$ in two equal right-angled triangles, $\Delta OAC$ and $\Delta OBC$.

Therefore, the area of the quadrilateral $OACB$ is given as,

$\Rightarrow 2\left(areaoftriangleOAC\right)=2\times \left(1/2\right)OA\times AC$

$=\sqrt{0+0+2g\times 0+2f\times 0+c}\sqrt{g^2+f^2-c}$

$=\sqrt{c(g^2+f^2-c)}$