## Precalculus: Mathematics for Calculus, 7th Edition

$(x-2)^{2}+(y+1)^{2}=9$
Center $(2,-1);$ radius $3$ The equation of a circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is its radius. Both the center and the radius of the circle are known. Substitute them into the formula to obtain its equation: $(x-h)^{2}+(y-k)^{2}=r^{2}$ $(x-2)^{2}+[y-(-1)]^{2}=(3)^{2}$ $(x-2)^{2}+(y+1)^{2}=9$