## Calculus: Early Transcendentals 8th Edition

$(x+1)^{2}+(y-5)^{2}=130$
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 $(h,k)=(-1,5)$ and $(x,y)=(-4, -6)$ Then $(-4+1)^{2}+(-6-5)^{2}=r^{2}$ $r^{2}=130$ Hence, $(x+1)^{2}+(y-5)^{2}=130$