Using spherical coordinates, is there an equation of a sphere not centered at the origin? If so what is it? I am a high school teacher teaching Calculus for the first time actually, I am teaching Multivariable Calculus (Calculus 3). Its been a solid 15 years since I took Calculus 3. During a discussion of the text my class and I came up with a question none of us could answer.

Each of the textbooks given examples of the spherical coordinates (of a sphere) is
centered at the origin. If the center of a given sphere is not at origin,
are there any changes in the spherical coordinates? If there is, how do I
manipulate that?

Now I am not certain if this is even considered, our text says it will not in rectangular coordinates, so I assume it won't in spherical. I am very rusty on this topic and would love additional insights or possible resources.
 A: Let’s consider the case of polar coordinates in the plane, which are usually given by the formulas
$$
x= r\cos \theta, \qquad y=r\sin \theta.$$
These coordinates are adapted to the circle of radius 1 with centre in the origin. If we wanted to prescribe a coordinate system adapted to the circle with centre $(x_0, y_0)$ and radius $L$, we should write
$$
x= x_0+Lr\cos \theta, \qquad y=y_0+ Lr\sin \theta.$$
The point at $r=0$ is now $(x_0, y_0)$. When $r=1$ and $\theta$ runs from $0$ to $2\pi$, the coordinates trace the circle with centre $(x_0, y_0)$ and radius $L$.
A: You can define "shifted" spherical coordinates. If the sphere's center is at $\mathbf r_0:=(x_0,y_0,z_0)$, then for $\mathbf r:=(x,y,z)$ define
$$\begin{align}
\rho(\mathbf r)&:=\vert \mathbf r-\mathbf r_0\vert\\
\theta(\mathbf r)&:=\cos^{-1}\left(\frac{z-z_0}{\vert \mathbf r-\mathbf r_0\vert}\right)\\
\varphi(\mathbf r)&:=\operatorname{atan2}(y-y_0,x-x_0)
\end{align}$$
It's really the same as spherical coordinates, but instead of using the vector pointing from the origin to a point, you use the vector pointing from the shifted center of the sphere to a point to calculate its shifted spherical coordinates. Or for the lazy: just pick the shifted center of the sphere as the origin to get the usual spherical coordinates.
If you want to describe a shifted sphere in the original spherical coordinates, though, that's going to be tough. The strength of a spherical coordinate system is that it makes descriptions of objects easier the more symmetries they have when rotated along an axis that goes through the origin. Like a sphere is perfectly easy to describe because it can be rotated along any axis through the origin. Cylinders are harder to describe in spherical coordinates, but still reasonably easy, because they have at least one axis through the origin along which they can be rotated. But a shifted sphere has no symmetries with respect to the origin, so its description in the usual spherical coordinates will be a mess.
