If two vertices of a quadrilateral are the centres of the circles
\[
S\equiv x^{2}+y^{2}-2x-2y-2=0
\]
and
\[
S^{\prime}\equiv x^{2}+y^{2}-6x-6y+14=0
\]
and the other two vertices of that quadrilateral are the points of intersection of these two circles, then the area of the quadrilateral is:
Show Hint
For circles having equal radii, the common chord is always perpendicular to the line joining centres. Area of the quadrilateral formed by centres and intersection points can be obtained using perpendicular diagonals.