jmc

Completing the square on $x^2+y^2-12x+31=0,$ we get $$(x-6{)}^2+y^2=5.$$Thus, the center of the circle is $(6,0),$ and its radius is $\sqrt{5}.$

Note that the parabola $y^2=4x$ opens to the right. Let $2t$ be the $y$-coordinate of $B.$ Then $$x=\frac{y^2}{4}=\frac{(2t{)}^2}{4}=t^2,$$so $B=(t^2,2t).$

Let $C=(6,0),$ the center of the circle.



By the Triangle Inequality, $AB+AC\ge BC,$ so $$AB\ge BC-AC.$$Since $A$ is a point on the circle, $AC=\sqrt{5},$ so $$AB\ge BC-\sqrt{5}.$$So, we try to minimize $BC.$

We have that $$\begin{array}{rl}B{C}^2& =(t^2-6{)}^2+(2t{)}^2\\ & =t^4-12t^2+36+4t^2\\ & =t^4-8t^2+36\\ & =(t^2-4{)}^2+20\\ & \ge 20,\end{array}$$so $BC\ge \sqrt{20}=2\sqrt{5}.$ Then $AB\ge 2\sqrt{5}-\sqrt{5}=\sqrt{5}.$

Equality occurs when $A=(5,2)$ and $B=(4,4),$ so the smallest possible distance $AB$ is $\boxed{\sqrt{5}}.$