Let S be the focus of the parabola y^ 2 =8 x and let P Q be the...

Given parabola $y^{2}=8 x and circle$x^{2}+y^{2}-2 x-4 y=0$pass through the origin$\therefore$One end of common chord PQ is origin. Say$\mathrm{P}(0,0)$Let$\mathrm{Q}$be the point \left(2 t^{2}, 4 t\right)$, then it will satisfy the equation of circle. $\therefore \quad 4 t^{4}+16 t^{2}-4 t^{2}-16 t=0 \Rightarrow t^{4}+3 t^{2}-4 t=0 \Rightarrow t\left(t^{3}+3 t-4\right)=0 \Rightarrow t(t-1)\left(t^{2}+t-4\right)=0 \Rightarrow t=0$ or 1 For $t=0$, we get point $P$, therefore $t=1$ gives point $Q$ as $(2,4)$. We also observe here that $\mathrm{P}(0,0)$ and $\mathrm{Q}(2,4)$ are end points of diameter of the given circle and focus of the parabola is the point $S(2,0)$. $\therefore \quad$ area $(\Delta \mathrm{PQS})=\frac{1}{2} \times P S \times Q S=\frac{1}{2} \times 2 \times 4=4$