If the common tangents to the parabola, ${\mathrm{x}}^2=4\mathrm{y}$ and the circle, ${\mathrm{x}}^2+{\mathrm{y}}^2=4$ intersect at the point P, then the distance of P from the origin, is

${\mathrm{x}}^2=4\mathrm{y} -\left(1\right) {\mathrm{x}}^2+{\mathrm{y}}^2=4 -\left(2\right)$

$\mathrm{y}=\mathrm{mx}-{\mathrm{m}}^2$ it is tangent of (2) then

   $$\begin{gathered} \mathrm{r}=\mathrm{d} \\ \Rightarrow 2=\frac{{\mathrm{m}}^2}{\sqrt{1+{\mathrm{m}}^2}} \\ \Rightarrow {\mathrm{m}}^4-4{\mathrm{m}}^2-4=0 \\ \Rightarrow {\left({\mathrm{m}}^2-2\right)}^2=6 \\ \Rightarrow {\mathrm{m}}^2-2=\pm 2\sqrt{2} \\ \Rightarrow {\mathrm{m}}^2=2\pm 2\sqrt{2} \\ \Rightarrow {\mathrm{m}}^2=2+2\sqrt{2} \\ Now \sin (90-\mathrm{\theta })=\frac{2}{\mathrm{d}} \\ \Rightarrow \mathrm{cos\theta }=\frac{2}{\mathrm{d}} \\ \Rightarrow \mathrm{d}=2\mathrm{sec\theta } \\ \Rightarrow {\mathrm{d}}^2=4{\mathrm{ec}}^2\mathrm{\theta } \\ = 4(1+{\tan }^2\mathrm{\theta }) \\ =4(3+2\sqrt{2}) \\ \mathrm{d}=2\sqrt{3+2\sqrt{2}}=2\left(\sqrt{2}+1\right) \end{gathered}$$