## Precalculus (6th Edition)

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.