## 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 $r^2=100$. Since the radius $r$ represents a distance, then it must be non-negative. This means $r^2$ must also be non-negative. Therefore, the given equation has no graph. There are no points on the graph of the given equation.