How to shift the circular hole of Dupin cyclide while keeping the curve circular without deformation?

Question

I have been interested in studying parametric functions. But when I dealt with Dupin cyclide, I found it difficult to shift the circular hole, for example, if I wanted to shift it towards the x-axis or the y-axis. There is always a deformation of Dupin's torus and it is close to the shape of the ellipse and it always needs adjustment. Is there a better way to do that.

\begin{align} x(u,v) &= \frac{d(c - a\cos u \cos v) + b^2 \cos u}{a - c \cos u \cos v}, \\ y(u,v) &= \frac{b\sin u (a-d\cos v)}{a - c \cos u \cos v},\\ z(u,v) &= \frac{b\sin v (c\cos u - d)}{a - c \cos u \cos v}. \end{align}

Answer 1

Try to translate the focal conics

• The first conic is focus-origin:

$$A= \left( \frac{(1+e)\cos u}{1+e\cos u}, \frac{(1+e)\sin u}{1+e\cos u}, 0 \right)$$

• The second conic is vertex-origin:

$$B= \left( \frac{e(1+\cos v)}{e-\cos v}, 0, \frac{(1+e)\sin v}{e-\cos v} \right) $$

• Divide $AB$ with the ratio $m:n$ gives the Dupin cyclide where

\begin{align} m &= \frac{e(1-\cos u)}{1+e\cos u}+k \\ n &= \frac{1+e}{e-\cos v}-k \end{align}

• For the circular hole centres at the origin,

$$k,e \in (0,1)$$

• For a torus,

$$e=0$$

• For a parabolic Dupin cyclide,

$$e=1$$

• Please also refer to Wikipedia article here.