The common tangents to the circle $x^2+y^2=2$ and the parabola $y^2=8x$ touch the circle at the points $P$,$Q$ and the parabola at the points $R$,$S$. Then the area of the quadrilateral $PQRS$ is

The equation of any tangent to the parabola can be considered as

$y=mx+\frac{a}{m}=mx+\frac{2}{m}$

i.e. $m^{2}x-my+2=0$

As we know that the length of the perpendicular drawn from the center to the tangent to the circle is equal to the radius of a circle.

Thus, $\frac{2}{\sqrt{m^4+m^2}}=\sqrt{2}$

$\begin{array}{l}\Rightarrow m^4+m^2=2\\ \Rightarrow m^4+m^2-2=0\\ \Rightarrow \left(m^2+2\right)\left(m^2-1\right)=0\\ \Rightarrow m=1\end{array}$

Hence, the equation of the tangents are 

$y=x+2$, $y=-x-2$

Therefore, the points $P$,$Q$ are $(-1,1)$, $(-1,-1)$ and $R$,$S$ are $(2,4)$ and $(2,-4)$ respectively.

Thus, the area of the quadrilateral $PQRS$

$=\frac{1}{2}\times (2+8)\times 3=15$ square units