Answer
A circle when $m>0$
A point when $m=0$
Nonexistent when $m<0$
Work Step by Step
$$(x-h)^2+(y-k)^2=m$$
Here if $m$ is negative, then there are no real solutions for ordered pair $(x,y)$.
It means such a circle does not exist.
If $m$ is zero, then we have a point that is ordered pair $(h,k)$.
If $m$ is positive, then we have a circle with center at $(h,k)$ and radius $\sqrt{m}$.
