## Precalculus (10th Edition)

$(x-4)^2+(y+1)^2=3^2.$
The standard equation for a circle with radius $r$ and centre $(h,k)$ is: $(x-h)^2+(y-k)^2=r^2$. Hence our equation: $(x-4)^2+(y-(-1))^2=3^2$, which is: $(x-4)^2+(y+1)^2=3^2.$