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 $$PC_1QC_2$$ is:

The two circles are

$$x^{2}+y^{2}-2x-2y-2=0$$ and $$x^{2}+y^{2}-6x-6y+14=0.$$

Write each in centre-radius form by completing squares.

First circle:
$$(x^{2}-2x)+(y^{2}-2y)=2$$
$$(x-1)^{2}-1+(y-1)^{2}-1=2$$
$$(x-1)^{2}+(y-1)^{2}=4.$$
Hence $$C_{1}(1,1), \; r_{1}=2.$$

Second circle:
$$(x^{2}-6x)+(y^{2}-6y)=-14$$
$$(x-3)^{2}-9+(y-3)^{2}-9=-14$$
$$(x-3)^{2}+(y-3)^{2}=4.$$
Hence $$C_{2}(3,3), \; r_{2}=2.$$

Subtracting the two equations to obtain the common chord:

$$(x^{2}+y^{2}-6x-6y+14)-(x^{2}+y^{2}-2x-2y-2)=0$$
$$-4x-4y+16=0 \;\Longrightarrow\; x+y=4.$$

This line is the radical axis and therefore contains the intersection points $$P,Q.$$

Put $$y=4-x$$ in the first circle:

$$x^{2}+(4-x)^{2}-2x-2(4-x)-2=0$$
$$x^{2}+x^{2}-8x+16-2x-8+2x-2=0$$
$$2x^{2}-8x+6=0$$
$$x^{2}-4x+3=0$$
$$(x-1)(x-3)=0 \;\Longrightarrow\; x=1 \text{ or } 3.$$

Hence the intersection points are

$$P(1,3), \qquad Q(3,1).$$

The required quadrilateral is $$P\,C_{1}\,Q\,C_{2}.$$ List its vertices in order:
$$P(1,3),\; C_{1}(1,1),\; Q(3,1),\; C_{2}(3,3).$$

Observe that
$$P \rightarrow C_{1}$$ is vertical of length $$2,$$
$$C_{1} \rightarrow Q$$ is horizontal of length $$2,$$
$$Q \rightarrow C_{2}$$ is vertical of length $$2,$$
$$C_{2} \rightarrow P$$ is horizontal of length $$2.$$
Thus the quadrilateral is a rectangle of side $$2 \times 2.$$

Area of rectangle $$= 2 \times 2 = 4 \text{ square units}.$$

Therefore, the area of $$P\,C_{1}\,Q\,C_{2}$$ is $$4.$$

Answer : Option B