Intersection of two quadrics that have a common inscribed sphere

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.

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.