find the shortest distance between y^2=4x and x^2+y^2-12x+31=0.
The firsdt equation describes a parabola opening to the right.
The second equation describes a circle around C(6, 0) with $\displaystyle r = \sqrt{5}$.
Enlarge the circle around C until it touches the parabola. The difference between the larger radius and the original radius is the shortest distance between the parabola and the circle:
$\displaystyle p: y^2=4x$
$\displaystyle c: (x-6)^2+y^2=r^2$ New circle with variable radius.
Point of intersection:
$\displaystyle (x-6)^2+4x=r^2$ Solve for x:
$\displaystyle x=4 \pm \sqrt{r^2-20}$
There is only one point of intersection, that means the 2 curves are tangent, if
$\displaystyle r = \sqrt{20} = 2\sqrt{5}$
The shortest distance is therefore:
$\displaystyle d_{min} = 2\sqrt{5}-\sqrt{5} = \sqrt{5}$