Torus and inversion yielding a given cyclide Let $\mathcal{C}$ be a ring cyclide with symmetry plane $\{z=0\}$ and parameters $a$, $c$, $\mu$ as in this picture:

How to find a torus and an inversion such that $\mathcal{C}$ is the image of this torus by this inversion?
 A: An answer is given in Garnier & al's paper Computation of Yvon-Villarceau circles on Dupin cyclides and construction of circular edge right triangles on tori and Dupin cyclides.
Let $k>0$ and define
$$
b = \sqrt{a^2-c^2},
$$
$$
r = \frac{kc^2(\mu-c)}{\bigl((a+c)(\mu-c)+b\sqrt{\mu^2-c^2}\bigr)\bigl((a-c)(\mu-c)+b\sqrt{\mu^2-c^2}\bigr)},
$$
$$
R = \frac{kc^2(a-c)}{\bigl((a-c)(\mu+c)+b\sqrt{\mu^2-c^2}\bigr)\bigl((a-c)(\mu-c)+b\sqrt{\mu^2-c^2}\bigr)},
$$
$$
\omega_2 = \frac{a\mu+b\sqrt{\mu^2-c^2}}{c},
$$
$$
\omega_{\mathcal{T}} = \omega_2 - \frac{kb^2(\omega_2-c)}{\bigl((a-c)(\mu+\omega_2)-b^2\bigr)\bigl((a+c)(\omega_2-c)+b^2\bigr)}.
$$
Then the inversion of center $(\omega_2,0,0)$ and ratio $k$ maps the torus centered at $(\omega_{\mathcal{T}},0,0)$ with major radius $R$, minor radius $r$, and symmetry plane $\{z=0\}$ to the cyclide $\mathcal{C}$ of implicit equation 
$$
{(x^2+y^2+z^2-\mu^2-b^2)}^2 - 4{(cx-a\mu)}^2 -4b^2z^2 = 0.
$$
EDIT
The above formula for $\omega_{\mathcal{T}}$ is wrong. With the notations of the paper, the formula $\omega_{\mathcal{T}} = \frac{b'_1+b'_2}{2}$ is correct. 
