## Precalculus (6th Edition)

$(4, -7)$
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: $(x-4)^2+(y-(-7))^2=3^2$ Thus, the center of the circle is at $(4, -7)$.