Sketching the surface of $z^2+{(\sqrt{(x^2+y^2)}-R)}^2-r^2=0$

Question

I'm having a problem visualizing the surface $F^{-1}({0})$ where $$F(x,y,z)=z^2+(\sqrt{x^2+y^2}-R)^2-r^2$$ with $0<r<R$ .

I have gotten the formula to the point $$z^2+x^2+y^2-2R\sqrt{x^2+y^2}=-R^2-r^2$$ but can't really seem to recognize any of the usual sphere/cylinder formulas.

I'm thinking this could be a sphere but the two different radii(i'm assuming this is what the r and R are) are throwing me off. Does anybody have ideas on how this particular surface could be sketched?

EDIT: Wolfram Mathrworld has an equation for the torus (http://mathworld.wolfram.com/Torus.html) which looks quite a bit like what I'm working with here.

Answer 1

This surface is a torus with eccentric dimension at z=0 equalling R for tube centre and tube radius r.

In the equation of the circle

$$z^2 +(x-R)^2-r^2=0$$

you have replaced $ x $ by $ \sqrt{x^2+y^2} $

which is how we can rotate the circle around $z$ axis to result in a torus:

$$z^2+{(\sqrt{ x^2+y^2}-R)}^2-r^2=0$$

Answer 2

I realize the formula I've reached was not as helpful as the initial one for F. This surface is a torus with outer radius R and inner radius r.