I need to parameterize the surface $$S=\{(x,y,z) \in \mathbb{R}^3 \colon (\sqrt{x^2 +y^2} -3 )^2+z^2 = 1\}.$$ My hint is that $S$ is a torus. I barely know where to begin. I have some idea on perhaps taking a circle and "lifting" it from the $xy$-plane to $z =$ radius of the torus and then let another circle centered on the first circle and orthogonal to the $xy$-plane and with the same radius as the torus and then "spin" it a full revolution around the $z$-axis but I can't make this into proper math and I'm not sure how to make it correspond to $S$.
I think I need two parameters: one for generating the circle orthogonal to the $xy$-plane and another for determine it's position around the $z$-axis but I'm stuck.
Spelling out what is already indicated by Did: Setting $r=\sqrt{x^2+y^2}$ the equation reads $(r-3)^2+z^2=1$, a circle of radius 1 centered at $(3,0)$ in the $(r,z)$ (half) plane. Indeed in the $(x,y,z)$ coordinates it is a torus (make a drawing). For the parametrization, set $r=(3+\cos s)$ and $z=\sin s$. And then $$ x=(3+\cos s) \cos t, \ \ y=(3+\cos s) \sin t, \ \ z=\sin s.$$