How to parametrize a Villarceau Circle? Basically the problem is to find a parametrization of a villarceau circle on a torus.  the hint is to try to find a bitangent plane to the torus and use the parametrization of the torus to find a parametrization of the circle. 
So in my attempt to solve this problem I took a torus with R, r being the distance from the center of the torus to the center of the tube, and the radius of the tube respectively. and R>r.  i tried to simplify my drawing by fixing the y axis at 0 which leaves you with two circles with a distance of 2R between their centers and I tried to find the equation for the line that is tangent to two points I found a diagonal line going from the inner top side of the left circle to the inner bottom side of the right circle(i wish i knew how to draw the picture here, but ill try to describe it as well as possible).  i ended up with a ratio  $z/x=-r/\sqrt{R^2-r^2}$ and $z/x=+r/\sqrt{R^2-r^2}$ giving the equation of a line(assuming we take the negative r) $z\sqrt{R^2-r^2}+xr=0$
my problem is going from here to an equation of a plane and once i get said plane, how to parametrize a circle on it so that it is on the surface of the torus.  any help is appreciated. thanks!
 A: Here is a brief outline.  You will need to fill in the details.
The torus with major radius $R$ and minor radius $r$ may be parameterized by
$$\displaylines{
P : [0,2\pi) \times [0, 2\pi) \longrightarrow \mathbb R^3 \\
P(\theta,\varphi) = (x,y,z) = \left((R + r \cos \varphi) \cos \theta , (R 
+ r \cos 
\varphi) \sin \theta, r \sin \varphi\right). \\
}$$
Now consider the rotation matrix 
$$M = {1 \over R}\begin{bmatrix}
\sqrt{R^2-r^2}&0&r \\
0&R&0 \\
-r&0&\sqrt{R^2-r^2}  \end{bmatrix}$$
which rotates the torus about an angle $\psi = \sin^{-1} (r/R)$ in the 
$xz$-plane.  We are interested in
$$P(M(x,y,z)) = \begin{bmatrix}
x' \\ y' \\ z'
\end{bmatrix} = 
\begin{bmatrix}
(R + r \cos\varphi)\cos\theta\cos\psi + r \sin\varphi\sin\psi \\
(R + r \cos\varphi)\sin\theta \\
-(R + r \cos\varphi)\cos\theta\sin\psi + r \sin\varphi\cos\psi 
\end{bmatrix},$$
and setting $z' = 0$ implies
$$(R + r \cos\varphi)\cos\theta = {r \sin\varphi\cos\psi \over \sin\psi}.$$
Hence $x' = R \sin\varphi$, and $(y')^2 = R \cos\varphi + r$.
therefore,
$$(x',y') = (R\sin\varphi, \pm(r + R\cos\varphi))$$
is a parameterization of the desired curve.
