## Calculus: Early Transcendentals 8th Edition

$(x+3)^2+(y-2)^2+(z-5)^2=4^2$ The sphere intersects the yz-plane at $(y-2)^2+(z-5)^2=4^2-3^2=7$
The equation can be found using the formula for a sphere: $(x-h)^2+(y-k)^2+(z-l)^2=r^2$ in which the point $(h,k,l)$ is the center of the sphere, and $r$ is the radius of the sphere. The intersection with the yz-plane can be determined by setting $x=0$ $(0+3)^2+(y-2)^2+(z-5)^2=4^2$ $(3)^2+(y-2)^2+(z-5)^2=4^2$ $(y-2)^2+(z-5)^2=4^2-3^2=7$