Distance Formula in Cartesian Plane

Q. Let $C_1$ and $C_2$ be the centres of the circles $x^2+y^2-2x-2y-2=0$ and $x^2+y^2-6x-6y+14=0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $P{C}_{1}Q{C}_2$ is :

1. $4$
2. $6$
3. $8$
4. $9$