Let 'P' be the point on the parabola ${\mathrm{y}}^2=4\mathrm{x}$ which is the shortest distance from the centre 'S' of the circle 

${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-16\mathrm{y}+64=0$ then SP =

$$\begin{gathered} \mathrm{Given} \mathrm{parabola} {\mathrm{y}}^2=4\mathrm{x} \\ \Rightarrow \mathrm{a}=1 \\ \mathrm{Given} \mathrm{circle} {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-16\mathrm{y}+64=0 \\ centre\left(\mathrm{S}\right)=(2,8) \end{gathered}$$

shortest distance  lies along common normal

Normal to parabola  is  $\mathrm{y}=-\mathrm{xt}+2\mathrm{t}+{\mathrm{t}}^3$

Since it passes through (2,8) then

$$\begin{gathered} \Rightarrow 8=-2\mathrm{t}+2\mathrm{t}+{\mathrm{t}}^3 \\ \Rightarrow \mathrm{t}=2 \\ Now \mathrm{P}=({\mathrm{at}}^2,2\mathrm{at})=(4,4) \\ \therefore \mathrm{SP}=\sqrt{4+16}=2\sqrt{5} \end{gathered}$$