# Locus of points equidistant to three spheres

Suppose we have three disjoint spheres in plain ordinary 3D space, with three different radii. I want to know the locus $L$ of points that are equidistant from these three spheres.

Partial answers: In 2D, the locus of points equidistant from two circles is a hyperbola. Therefore, in 3D, the locus of points equidistant from two spheres is (one half of) a hyperboloid of two sheets. So, the locus $L$ that I'm seeking is the intersection of two such hyperboloids. Based on some experiments, it seems that $L$ is a planar curve, and therefore a conic section. If that's true, then the proof ought to be very simple, but I don't see it.

• I don't see why the curve would be a conic section (or planar, for that matter). Jun 30, 2017 at 7:54
• @IvanNeretin -- well in the few cases I've tried, it is a conic. Do you have an example where it isn't? Jun 30, 2017 at 7:59
• Nope. So looks like I'm missing something, and there is a reason, after all... Jun 30, 2017 at 9:34

Consider the equation of the hyperboloid between $S_1$ and $S_2$, $$\|\mathbf x-\mathbf p_1\|-r_1=\|\mathbf x-\mathbf p_2\|-r_2.$$ With a little algebra, we can express this as $$2(r_2-r_1)\|\mathbf x-\mathbf p_1\|=2(\mathbf p_2-\mathbf p_1)\cdot\mathbf x+\underbrace{\|\mathbf p_1\|^2-\|\mathbf p_2\|^2+(r_2-r_1)^2}_{c_2}.$$ Similarly, the hyperboloid between $S_1$ and $S_3$ is of the form $$2(r_3-r_1)\|\mathbf x-\mathbf p_1\|=2(\mathbf p_3-\mathbf p_1)\cdot \mathbf x+c_3.$$ Cancelling terms proportional to $\|\mathbf x-\mathbf p_1\|$ from the two equations, we obtain \begin{align} 0 &= (r_3-r_1)\bigl(2(\mathbf p_2-\mathbf p_1)\cdot\mathbf x+c_2\bigr) - (r_2-r_1)\bigl(2(\mathbf p_3-\mathbf p_1)\cdot\mathbf x+c_3\bigr) \\ &= 2\bigl((r_2-r_3)\mathbf p_1 + (r_3-r_1)\mathbf p_2 + (r_1-r_2)\mathbf p_3\bigr)\cdot\mathbf x + \text{const}, \end{align} which is the equation of a plane orthogonal to $(r_2-r_3)\mathbf p_1 + (r_3-r_1)\mathbf p_2 + (r_1-r_2)\mathbf p_3$.

Two approaches; the first is essentially the same as Rahul's, but slightly tidier (in my opinion).

Without loss of generality, we can shrink the radii of all three spheres until the smallest becomes zero. This will not change the equidistant locus. Also, we may assume that the zero-radius sphere, $S_0$, is located at the origin. Call the other two spheres $S_p$ (with center at $\mathbf{p}$ and radius $a$) and $S_q$ (with center at $\mathbf{q}$ and radius $b$).

At points $\mathbf{x}$ that are equidistant from $S_0$ and $S_p$, we have $$\|\mathbf{x}\| = \|\mathbf{x} - \mathbf{p}\| - a$$ After some algebra, this gives us $$2a\|\mathbf{x}\| = 2 \mathbf{p} \cdot \mathbf{x} - \|\mathbf{p}\|^2 + a^2$$ Similarly, points that are equidistant from $S_0$ and $S_q$ satisfy $$2b\|\mathbf{x}\| = 2 \mathbf{q} \cdot \mathbf{x} - \|\mathbf{q}\|^2 + b^2$$ Eliminating $\|\mathbf{x}\|$ from these last two equations, we get $$a( 2 \mathbf{q} \cdot \mathbf{x} - \|\mathbf{q}\|^2 + b^2 ) = b( 2 \mathbf{p} \cdot \mathbf{x} - \|\mathbf{p}\|^2 + a^2 )$$ and re-arranging gives $$2\mathbf{x} \cdot (a\mathbf{q} - b\mathbf{p}) = a\|\mathbf{q}\|^2 - b\|\mathbf{p}\|^2 +a^2b - ab^2$$ This is the equation of a plane whose normal is in the direction $a\mathbf{q} - b\mathbf{p}$.

A Different Approach
The locus is the path of sphere (of varying radius) which is moving so as to remain tangent to the given three spheres. The envelope of such a moving sphere is a Dupin cyclide. The centerline of a Dupin cyclide is a conic. Details in this report.

• I get sign-reversed left-hand sides for the 2nd and 3rd eqns, but that does not affect the conclusion. Nice! Jul 1, 2017 at 8:45
• I should have noticed that both sides of my equations were squares... /facepalm
– user856
Jul 1, 2017 at 11:04
• @Rahul -- the algebra (both yours and mine) involves a lot of squaring of things, which might introduce extraneous solutions. I thought your squared terms were intended to control/mitigate this. Jul 2, 2017 at 0:22