If a circle with its centre at the focus of the parabola  $y^2=2px$. Is such that it touches the directrix of the parabola, then a point of intersection of the circle and the parabola is 

Focus= $(\frac{p}{2},0)=centreofole$
Radius of ole is “P”
Equation of ole :  ${(x-\frac{p}{2})}^2+{\left(y\right)}^2=p^2 \mathrm{.........}\left(1\right)$
 $e{q}^{n}ofparabola:y^2=2px \mathrm{........}\left(2\right)$
Solving (1) & (2)
$\begin{array}{l}{(x-\frac{p}{2})}^2+2px=p^2\\ \Rightarrow 4x^2+4px=3p^2\\ \Rightarrow (2x+3p)(2x-p)=0\\ \Rightarrow x=\frac{p}{2},x=-\frac{3p}{2}\end{array}$

$When \Rightarrow x=\frac{p}{2}\Rightarrow y^2=p^2$

When  $x=-\frac{3p}{2}\Rightarrow y^2=-\frac{6p^2}{2} \left(rejected\right)$

$\therefore ptof\text{\hspace{0.17em}intersection\hspace{0.17em}are }(\frac{p}{2},p)and(\frac{p}{2},-p)$