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 $OABC$ is 

Since $OA=OB$ and $CA=CB$, the diagonal $OC$ divides the quadrilateral OACB in two equal right-angled triangles, $OAC$ and $OBC$  Therefore, the area of the quadrilateral $OACB$ is

2(area of triangle $OAC$) $=2\times (1/2) OA\times AC$

$\begin{array}{l}=\sqrt{0+0+2g\times 0+2f\times 0+c}\sqrt{g^2+f^2-c}\\ =\sqrt{c\left(g^2+f^2-c\right)}\end{array}$