Let $C_{1}and C_2$ be the centers 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
  

Let $S_1\equiv x^2+y^2-2x-2y-2=0$

$\begin{array}{r}{S}_2\equiv x^2+y^2-6x-6y+14=0\\ \therefore C_1\equiv (1,1)\text{ and }{C}_2\equiv (3,3)\end{array}$

Radius of $S_1,P{C}_1=\sqrt{1+1+2}=2$

Radius of $S_2,P{C}_2=\sqrt{9+9-14}=2$

$\therefore P{C}_1=Q{C}_1=P{C}_2=Q{C}_2=2$

Now, $2g_1{g}_2+2f_1{f}_2=2\times 3+2\times 3=6+6=12$ and 

$c_1+c_2=14-2=12$

Here, $2g_1{g}_2+2f_1{f}_2=c_1+c_2$

$\therefore$Both circles are orthogonal. So, $P{C}_{1}Q{C}_2$ is a square.

$\therefore$ Area of $P{C}_{1}Q{C}_2=2\times 2=4\text{ sq. units }$.