#### Answer

one point only

#### Work Step by Step

RECALL:
The circle $(x-h)^2+(y-k)^2=r^2$ has its center at $(h, k)$ and a radius of $r$ units.
The given equation has its "center" at $(0, 0)$ and has $r=0$.
Since the radius is zero, then the graph must only be a point - $(0, 0)$,
Thus, the given equation has only one point on its graph.