Intersection of two quadrics that have a common inscribed sphere This is related to a question I asked here on math.stackexchange. It didn't receive an answer there (except for my answer), and my question here is a generalization of that one, anyway.
Suppose I have two quadric surfaces $Q1$ and $Q2$ in plain ordinary 3D space, $\mathbb{R}^3$, and they have a common inscribed tangent sphere $S$, as shown in the picture. 

In general, the intersection of two quadric surfaces is a nasty curve of degree 4. But in our case, because of the common tangency, I suspect that the intersection of $Q1$ and $Q2$ is just a pair of ellipses. Numerical experiments seem to suggest this, anyway. My question: is my conjecture true, and, if so, what is the proof?
My tags might well be wrong, so please feel free to edit.
 A: In fact, there is a far more general result. In Salmon's "Analytic Geometry of Three Dimensions", 4th edition, page 117, we find the following:

Two quadrics having plane contact with the same third quadric intersect each other in plane curves.  Proof: Obviously $U-L^2$ and $U-M^2$ have the planes $L-M$ and $L+M$ for their planes of intersection.

This is essentially the same proof given in Ivan's answer.
A: This is a nice observation about quadrics, I haven't seen it stated anywhere. It can be proved with the help of pencils of quadrics.
1) The curve of tangency is a circle.
Consider the pencil of quadrics spanned by $S$ and $Q_1$ (the set of quadrics whose equations are linear combinations of those for $S$ and $Q_1$). All quadrics of the pencil are tangent along the same curve, and somewhere in the pencil there is a degenerate quadric, which is a cylinder or a cone.
2) The quadric $Q_1$ can be described by an equation of the form
$$\|x\|^2 -r^2 + c_1 f_1^2 = 0,$$
where $r$ is the radius of the sphere.
Indeed, the pencil through $S$ and $Q_1$ contains the double plane through the circle of tangency. Let $f_1 = 0$ be the equation of this plane. The quadric $Q_1$ can be described by a linear combination of the equations for $S$ and the equation $f_1^2 = 0$ of the double plane.
Similarly, $Q_2$ has an equation of the form
$$\|x\|^2 - r^2 + c_2f_2^2=0.$$
Take the difference of the above equations. The intersection of $Q_1$ and $Q_2$ is contained in the union of two planes $\sqrt{c_1}f_1 \pm \sqrt{c_2}f_2 = 0$. Each of these planes intersects $Q_i$ along a conic.
