I am trying to study the curve determined by intersecting a torus with center (0,0,0), inner radius $a-r$ and outer radius $a+r \ (0<r<a)$ with the plane $x=a-r$. This intersection looks particularly interesting, since the resulting curve is a horizontal figure eight, given by the equation
$$z(y) = \pm \sqrt{r^2-\left( a - \sqrt{(a-r)^2+y^2} \right)^2}.$$
I can't do much with this equation, though, so I was at least trying to determine the slope of the lines tangent at (0,0) (in the $yz-$plane). So I chose the upper branch of $z(y)$ as a function, differentiated it and obtained
$$z'(y) = \dfrac{\dfrac{ay}{\sqrt{(a-r)^2+y^2}} -y}{\sqrt{r^2-\left(a-\sqrt{(a-r)^2+y^2}\right)^2}}.$$
Turns out, as expected, that inserting $y=0$ gives $\frac{0}{0}$, since there is an angle at that point. So I tried to use Bernoulli-L'Hôpital (considering only the limit $y\to 0+$) once or twice, but it looks like the nasty big square root in the denominator never goes away, so I never get a valid result and the equations become more and more complicated.
I also tried to insert values like $a=3$ and $r=1$ to get easier computations, but it does not help. Any suggestions?
About the slope, see below.
The intersection curve is a classical curve called a lemniscate of Bernoulli : see "slicing a torus" in (http://xahlee.info/SpecialPlaneCurves_dir/LemniscateOfBernoulli_dir/lemniscateOfBernoulli.html). More generally you get Cassinian ovals by vertical slicing of a torus (http://www.mathcurve.com/courbes2d.gb/cassini/cassini.shtml). See also (http://www.cgl.ucsf.edu/home/bic/torus/torus-slicing.pdf).
If you slice the torus by a non vertical plane passing through the origin with a convenient "slope", the locus is a pair of "Villarceau circles" (http://paulbourke.net/fun/toruscut/)
About the slope, instead of a cartesian equation, consider a description of the lemnicate by parametric equations or a polar equation (Parametrization of the lemniscate)