Sphere tangential to a plane and two spheres of different radius

Question

Let $\pi$ be a plane and $S_1, S_2$ two external spheres of different radius that belong to the same semi-space identified by $\pi$. Demonstrate that a sphere simultaneously tangential to $\pi,S_1,S_2$ exists. Is there an infinite number of such a sphere?

I came across this problem in an old test of linear algebra so i suppose that i should be able to solve it with the use of my knowledge (matrix, rank, norm, bilinear form, ecc). The only thing i have noticed is this:

Let $C_1$ and $R_1$ be the center and radius of $S_1$, $C_2$ and $R_2$ be the center and radius of $S_2$, $C$ and $R$ be the center and radius of the hipotetic sphere: $$d(C,C_1)=R+R_1 \quad d(C,C_2) = R+R_2 \quad d(\pi,C) = R$$

Unfortunately i have no idea of what should i do in this kind of problem and i haven't got a solution or a solved exercise of this kind. I would really appreciate any hint or suggestion on this problem. Thank you in advance!

Answer 1

ok a hint (even though the approach may not be to your liking):

Start out in the two dimensional case with two circles and a line.

Assume the circles have radii $r < R$ and the distance between the centers $c, C$ is $d=r+R+ a$ with $a>0$. Take any number $b> \frac{a}{2}$ (small letter $c$ corresponds to small letter $r$).

If you draw a circle of radius $S = R+b$ with center $C$ and another circle of radius $s=r+b$ around $c$ these two circles will intersect in two points, $p_1$ and $p_2$, say.

It is easy to see that the intersection of the lines from $c$ to $p_1$ and from $C$ to $p_1$ intersect in a point $Q$. The circle around $Q$ of radius $b$ is tangent to both the circles you started with (it will meet them in the intersection of the line from $c$ to $Q$ (from $C$ to $Q$) with the circle around $c$ ($C$)). A similar statement is true for $p_2$.

Now you can either let $b$ vary from $\frac{a}{2}$ to $\infty$ or try some elementary geometric reasoning to find $b$ such that the circle around $Q$ (which depends on $b$, of course) ist tangent to the line $\pi$.

A similar approach works in three dimensions, too, it's just a bit more complicated to write down the details.

Answer 2

Notice first of all that you considered only one of the possible cases, when the sphere is externally tangent to the two given spheres, but other cases are possible, if you allow the spheres to touch internally.

We can set up a coordinate system such that plane $\pi$ has equation $z=0$: if $C=(x,y,z)$ is the center of the tangent sphere, your third equation reduces then to $R=z$. Inserting this value of $R$ into the other two equations yields the equations of two quadrics, whose intersection will then give you the locus of $C$.

By subtracting the equations of those quadrics all quadratic terms vanish, and you end up with the equation of a plane. The locus of $C$ is then the intersection of this plane with either quadric, that is a conic section.