find the shortest distance between y^2=4x and x^2+y^2-12x+31=0.
The second equation describes a circle around C(6, 0) with .
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:
New circle with variable radius.
Point of intersection:
Solve for x:
There is only one point of intersection, that means the 2 curves are tangent, if
The shortest distance is therefore: