Minimum distance between the curves $y^2=4 x$ and $x^2+y^2-12 x+31=0$ is

Centre and radius of the given circle is $P(6,0)$ and $\sqrt{5}$, respectively. Now minimum distance between two curves always occurs along a line which normal to both the curves. Equation of normal to $y^2=4 x$ at $\left(t^2, 2 t\right)$ is $y=-t x+2 t+t^3$
If it is normal to circle also, then it must pass though $(6,0)$. $\therefore 0=t^3-4 t \Rightarrow t=0 \quad$ or $t= \pm 2$. $\begin{aligned} & \Rightarrow A(4,4) \text { and } C(4,-4) \\ & \Rightarrow P A=P C=\sqrt{20}=2 \sqrt{5} . \\ & \Rightarrow \text { Required minimum distance }=2 \sqrt{5}-\sqrt{5}=\sqrt{5}\end{aligned}$