## Calculus 8th Edition

$(x-3)^{2}+(y+1)^{2}=25$
An equation of the circle with center $(h,k)$ and radius $r$ is given as: $(x-h)^{2}+(y-k)^{2}=r^{2}$ Given: center,$(h,k)=(3,-1)$ and radius, $r=5$ Then $(x-3)^{2}+(y+1)^{2}=5^{2}$ Hence, $(x-3)^{2}+(y+1)^{2}=25$