find mini distance

The curve $y^2=4x$ is a parabola symmetrical about the $x$ -axis while $x^2+y^2-12x+31=0$ $\implies (x-6)^2 + y^2=5$ is a circle of radius $\sqrt 5$ and centered at $C (6,0)$ . The the distance between a point $P(x,y)$ on the parabola and the circle is the length $\overline{PC}$ minus the radius of circle $\sqrt 5$ or

$\begin{aligned} d & = \sqrt {(x-6)^2+(y-0)^2} - \sqrt 5 \\ & = \sqrt {(x-6)^2+y^2} - \sqrt 5 \\ & = \sqrt {x^2-12x+36+4x} - \sqrt 5 \\ & = \sqrt {x^2-8x+36} - \sqrt 5 \\ & = \sqrt {(x-4)^2+20} - \sqrt 5 \end{aligned}$

Therefore, $d$ is minimum when $x-4=0$ . That is $\min(d) = \sqrt{20}-\sqrt 5 = \sqrt 5 \approx \boxed{2.236}$ .