A circle drawn having centre at C(0,2) and passing through focus (S) of the parabola ${\mathrm{y}}^2=8\mathrm{x}$, if radius (CS) , intersects the parabola at point P, then

According to the given co-ordinates of C and co-ordinate of focus we can see if we plot the diameter through C and S the other end ( say Q ) lies on directrix of parabola . Since , the circle must pass through ‘R’ 
$\overline{\mathrm{CS}}$ Equationis x + y = 2
And parabola Equation ${\mathrm{y}}^2=8\mathrm{x}$

solving line and parabola
$\begin{array}{l}\Rightarrow \mathrm{x}-\text{ coordinate of ' }{\mathrm{P}}^{\mathrm{\prime }}\mathrm{is}(6-4\sqrt{2})\therefore \mathrm{P}=(6-4\sqrt{2},4\sqrt{2}-4)\\ \text{ Now }\overline{\mathrm{CP}}=\mathrm{r}-\mathrm{sp}=6\sqrt{2}-8\end{array}$
Slope of $\overline{\mathrm{CQ}}\times \text{ slope of}\text{ of }\mathrm{CR}=-1{\Rightarrow }^{\mathrm{\prime }}{\mathrm{C}}^{\mathrm{\prime }}$is also correct.