$\text{ The minimum distance between the curves }{y}^2=4x\text{ and }{x}^2+y^2-12x+31=0\text{ is }$

$\text{ Centre and radius of the given circle are }P(6,0)\text{ and }\sqrt{5}\text{ , respectively.}$

Now, the shortest distance always occurs along the common normal.

$\text{ Differentiating }{y}^2=4x\text{ with respect to x we get,}$

$\frac{dy}{dx}=\frac{2}{y}$

$\text{Then the slope of nonnal at point} A\left(y_1^2/4,y_1\right)\text{ is }-y_1/2$

$\text{ Also, from definition, the slope of }AP\text{ is given by }\frac{y_1-0}{\left(y_1^2/4\right)-6}=-\frac{y_1}{2}$

$\text{ i.e., }{y}_1=0\text{ or }{y}_1=\pm 4$

$\text{ Hence, the points are }O(0,0),A(4,4),\text{ and }C(4,-4)\text{ . }$

$\text{ The shortest distance is }AP-\sqrt{5}=\sqrt{20}-\sqrt{5}=\sqrt{5}\text{ . }$