Although it is known the solution to the first two questions, somebody may have different nice answers, so I include them:

  1. Given two circles in the plane, there is (at least) a line which is tangent to both of them.

  2. Given three spheres in the space, there is a plane which is tangent to all of them.

  3. In general, given $n$ n-spheres in the n-dimensional space, is there a hyperplane which is tangent to all of them?

  4. What other generalizations does this problem admit?

EDIT: As @Noam kindly remarked below, the existence of the tangent objects is not always true. I think that in #2 the hypothesis must be: one of the spheres not in the cone determined by the other two. In #1 the "cone" determined by one circle is the circle itself. So in #3 we need a suitable definition for the "cone" determined by $n-1$ n-spheres.

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    $\begingroup$ You might be interested in Soddy's hexlet and generalizations, for a sphere of infinite radius is a plane: en.wikipedia.org/wiki/Soddy%27s_hexlet $\endgroup$ – Joseph O'Rourke Jan 30 '12 at 17:57
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    $\begingroup$ Already #1 is not true in ${\bf R}^2$ if one of the circles is strictly inside the other... For #2 and beyond there are further counterexamples (imagine two large balls and a tiny ball floating at a suitable distance between them). In general (#3), the hyperplanes in ${\bf R}^n$ constitute an $n$-dimensional linear space, and tangency to a given hypersphere imposes a degree-$2$ condition, so there will be $2^n$ solutions counted with multiplicity, but some or all of them might exist only over ${\bf C}$. As for #4, The same is true with arbitrary quadrics in place of spheres. $\endgroup$ – Noam D. Elkies Jan 30 '12 at 19:01
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    $\begingroup$ Re #4: The ultimate generalization (or, at least, one of them) is, I guess, intersection theory and enumerative geometry. Apollonius' problem (I am sure this is somewhere in Wikipedia) is a classical application of this, and this is more or less your #1 $\endgroup$ – Mariano Suárez-Álvarez Jan 30 '12 at 19:44
  • $\begingroup$ When you say "cone" do you mean "convex hull"? $\endgroup$ – S. Carnahan Jan 31 '12 at 5:18
  • $\begingroup$ I think "cone" is appropriate. There are tangent planes which separate two spheres, and Noam's suggested example applies if the tiny sphere avoids the separating double cone. Gerhard "Ask Me About System Design" Paseman, 2012.01.30 $\endgroup$ – Gerhard Paseman Jan 31 '12 at 5:39

This answers expands on my comment on the original question.

1. Given two circles in the plane, there is (at least) a line which is tangent to both of them: this is not true unless we allow lines with complex coefficients (and even then there's an exception, see below). In the real plane, two circles have:

$\bullet$ no common tangents if one is contained in the other's interior,

$\bullet$ two common tangents if they intersect at two points, and

$\bullet$ four common tangents if they're disjoint and neither's interior contains the other.

At the boundaries between these cases, the circles can have 1 or 3 common tangents if they're tangent internally or externally. There's also a generate case where the two circles coincide and thus have infinitely many common tangents.

2. Given three spheres in the space, there is a plane which is tangent to all of them: again not true, and and with a wider variety of counterexamples, and a wider variety of degenerate cases where there are infinitely many common tangents. Aside of these degenerate examples, the maximum number of common tangent planes is $8$. (See below for how to compute them.) A simple configuration with infinitely many tangent planes is three identical spheres with collinear centers. More generally, three spheres with collinear centers that have one tangent plane have infinitely many, because we can rotate the plane about the line $l$ joining the centers. But make one sphere a bit larger or smaller about the same center, and there's no common tangent plane at all, and then this stays true if we move the centers a bit off $l$.

3. In general, given $n$ spheres in $n$-dimensional space, is there a hyperplane which is tangent to all of them? [The proposer wrote "$n$ n-spheres", but current terminology uses "n-sphere" for a sphere in ${\bf R}^{n+1}$.] Again, not necessarily; the number can range from $0$ to $2^n$, or be infinitely large, and there are various ways for the number to be zero, most simply when one sphere is contained in another's interior.

The hyperplanes in ${\bf R}^n$ constitute an $n$-dimensional space, more precisely a projective space ${\bf RP}^n$ with a point removed (the hyperplane at infinity, which is not relevant to us because it is not tangent to any sphere). The hyperplanes tangent to a given sphere constitute a degree-$2$ hypersurface in this projective space. Generically $n$ such surfaces in $n$-dimensional space meet in $2^n$ points, but it is possible for none of them to have real coordinates; and there are also special configurations that intersect in a positive-dimensional variety, most simply when they are linearly dependent (but this is far from the only reason, as noted in this mathoverflow answer).

4. What other generalizations does this problem admit? I wrote that one gets the same $2^n$ enumeration with $n$ arbitrary quadrics in place of spheres, but Mariano Suárez-Alvarez suggested a much more complete answer:

The ultimate generalization (or, at least, one of them) is, I guess, intersection theory and enumerative geometry.

Still, there are special features of the $n$-sphere problem that do not survive generalization even to $n$ quadrics. Most notably, the $2^n$ tangent planes can be given in closed form, each requiring only the extraction of a single square root. Suppose the spheres have centers $v_i \in {\bf R}^n$ and radii $r_i \in (0,\infty)$. A hyperplane $a \cdot x = b$ is tangent to the sphere $|x-v_i|=r$ iff $a \cdot v_i - b = \pm r_i |a|$. Now for each of $2^n$ choices of $\pm$ signs we get $n$ simultaneous linear equations in $n+1$ variables: the coordinates of $a$, and the length $|a|$. Generically there is a line of solutions, and then the additional condition that $|a|$ really be the length of the vector $a$ gives a quadratic equation on that line which usually has $2$ solutions, either real or complex conjugate. [This gives $2^n$ hyperplanes, not twice that number, because switching all the $\pm$ signs yields the same solution.] But there are also non-generic cases where the equations are dependent, and so give a higher-dimensional space of solutions (not necessarily defined over ${\bf R}$), or inconsistent, and so give no solutions at all. If I did this right, one of these happens iff the centers of the spheres lie on a linear subspace of dimension at most $n-2$. So for $n=2$ we're back to concentric circles, with infinitely many solutions if they coincide, and none otherwise, not even over ${\bf C}$. Likewise for $n=3$ we get the configurations of three spheres with collinear centers.

  • $\begingroup$ It'd be nice to be able to point to some standard reference where the case of $n$ general quadrics in projective $n$-space is treated using the standard tools of enumerative geometry. While it is not difficult (the hyperplanes tangent to a quadric form a quadric in the dual projective space, so we are counting the number of intersection points in the intersection of $n$ quadrics in the dual $P^n$: Bézout's theorem immediately gives us Noam's $2^n$), surely someone wrote this down in some book? $\endgroup$ – Mariano Suárez-Álvarez Jan 31 '12 at 6:56

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