Must three distinct spheres always intersect in exactly two points?

Question

Two spheres A, B intersect in a circle, obviously, so a 3rd sphere C intersecting both A and B does so in 2 different circles.

It seems to me that the circle of the AC intersection must intersect the circle of the BC intersection in exactly two points, but a proof — even an informal one — eludes me. I can't think of a way (in a proof) to force the two circles AC and BC to intersect at all.

Answer 1

A useful trick: if $f(x,y,z) = 0$ and $g(x,y,z) = 0$ are the equations of two spheres $S_1$ and $S_2$ with distinct centers, then $h = f - g$ is linear, and therefore $h = 0$ is the equation of a plane $P$. Further, if any two of $f, g, h$ vanish then so does the third, or in other words, the intersection of any two of $S_1, S_2, P$ is contained in the third, ie equals the intersection of all three.

Now, the intersection of two spheres is the intersection of a plane and a sphere, and if you restrict the equation of a sphere to a plane you get a possibly degenerate circle (ie could be a single point or empty). If you intersect three distinct spheres, then that's the same as intersecting two planes with a sphere (or one of the pairs has the same center and therefore empty intersection). If we first intersect the two planes, we can get a line, a plane, or an empty intersection (if they are parallel and not coincident). Now the intersection of that with a sphere has to be one of: a circle, two points, one point (if the line or plane is tangent to the sphere) or the empty set.

All of these are possible, and all but the two points case are possible with all the centers being collinear.

Answer 2

As stated, it is possible the three spheres A. don't intersect at all, B. they intersect in a single point, C. they intersect in a circle or D. they intersect in two points.

Example of A:
Denote each center as $C_i$, each radius as $r_i$ $C_1 = (-10, 0, 0), C_2 = (0, 0, 0), C_3 = (10, 0, 0), r_1= r_2=r_3 = 1$

Example of B:
$(x-1)^2+y^2+z^2=1, (x-2)^2+y^2+z^2=2^2, (x-3)^2+y^2+z^2=3^2$ They all have the origin as a point in common.

Example of C:
$(x-2)^2+y^2+z^2=3, (x+2)^2+y^2+z^2=3^2, (x-3)^2+y^2+z^2=r_1^2$ Note: Choose r_1 so that the intersection of the third circle coincides with the intersection of the first two.

Another example of A:
Use C, except increase the radius, so that any two pairs of spheres intersect but the three have no mutual intersection.

Example of D:
$(x-2)^2+y^2+z^2=3, (x+2)^2+y^2+z^2=3^2, x^2+(y-1)^2+z^2=r_2^2$

Thus it is shown by example that other possibilities for the intersections exist.

Answer 3

Two spheres of centers $C_1$ and $C_2$ and radii $r_1$ and $r_2$ , respectively, will

• Not intersect if $\| C_1 C_2 \| \gt r_1 + r_2 $ or if $ \| C_1 C_2 \| \lt | r_2 - r_1 | $

• Intersect at only one point if $ \| C_1 C_2 \| = | r_1 - r_2 | $ or if $ \| C_1 C_2 \| = r_1 + r_2 $

• Intersect in a circle otherwise, i.e. if $|r_2 - r_1| \lt \| C_1 C_2 \| \lt r_1 + r_2 $

Therefore, when we introduce the third sphere, there can be zero, one, or two intersections, or otherwise an infinite number of intersections that are the same circle as the circle of intersection of the first two spheres. In particular, we will have only one intersection if the first two spheres intersect at one point which lies on the third sphere, or if the third sphere is tangential to the circle of intersection of the first two spheres.

So, to summarize, there can be no intersections, one intersections, two intersections, or infinitely many intersections. In the last case, the intersection is a circle. We're of course assuming that the three spheres are distinct.