## 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 can be written as $(x-3)^2+(y-(-2))^2=5^2$ This circle has its center at $(3, -2)$ and has a radius of 5 units. Therefore, its graph is the one in Option C.