The parametric equations of the parabola $y^{2}=8 x$ are $x=2 t^{2}$ and $y=4 t$

and the given equation of circle is

$x^{2}+y^{2}-2 x-4 y=0$

On putting $x=2 t^{2}$ and $y=4$ t in circle

we get

$$

\begin{array}{l}

4 t^{4}+16 t^{2}-4 t^{2}-16 t=0 \\

\Rightarrow 4 t^{2}+12 t^{2}-16 t=0 \\

\Rightarrow 4 t\left(t^{3}+3 t-4\right)=0 \\

\Rightarrow t(t-1)\left(t^{2}+t+4\right)=0 \\

\Rightarrow t=0, t=1 \\

\quad\left[\because t^{2}+t+4 \neq 0\right]

\end{array}

$$

Thus the coordinates of points of intersection of the circle and the parabola are $\mathrm{Q}(0,0)$ and $\mathrm{P}(2,4) .$ Clearly these are diametrically opposite points on the circle.

The coordinates of the focus $\mathrm{S}$ of the parabola are (2,0) which lies on the circle.

$$

\begin{aligned}

\therefore \text { Area of } \Delta \mathrm{PQS} &=\frac{1}{2} \times \mathrm{QS} \times \mathrm{SP}=\frac{1}{2} \times 2 \times 4 \\

&=4 \mathrm{sq} . \text { units. }

\end{aligned}

$$