Answer
The equation of the sphere is $(x+3)^2+(y-2)^2+(z-5)^2=16$.
The intersection is circle in the yz-plane with center (0,2,5) and radius $\sqrt{7}.$
Work Step by Step
We are given the center of the sphere is $(-3,2,5)$ and its radius is 4. So the equation of the sphere is $$(x+3)^2+(y-2)^2+(z-5)^2=16.$$
Now, to find the intersection of our sphere with the yz-plane, we must find the set of points $(x,y,z)$ satisfying the equation of our sphere and the equation of the yz-plane.
The equation of the yz-plane is $x=0$. So this intersection is the set of all points $(x,y,z)$ such that
$$x=0$$ and $$(x+3)^2+(y-2)^2+(z-5)^2=16.$$
Note for points with zero x-coordinate, the equation of our sphere becomes
$$(0+3)^2+(y-2)^2+(z-5)^2=16
\\9+(y-2)^2+(z-5)^2=16\\(y-2)^2+(z-5)^2=7.$$
So our intersection is the set of all points $(x,y,z)$ such that
$$x=0$$ and $$(y-2)^2+(z-5)^2=7.$$
Hence our intersection forms a circle in the yz-plane with center (0,2,5) and radius $\sqrt{7}.$
