## Precalculus (6th Edition)

$\color{blue}{(x-3)^2+(y-6)^2=16}$
RECALL: The circle $(x-h)^2+(y-k)^2=r^2$ has its center at $(h, k)$ and a radius of $r$ units. Thus, the equation of a circle with center at $(3, 6)$ and a radius of $4$ is: $(x-3)^2+y-6)^2=4^2 \\\color{blue}{(x-3)^2+(y-6)^2=16}$