I want to find the planes tangent to three given circles in 3D space.

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I'm not sure how many solutions there are, in general. My guess is that there are 8. This suggests that we might have to find the roots of some polynomial of degree 8, which would be bad news.

A Google search for "plane tangent to three circles" yields exactly one result, which is this question. It was asked in 2011, and was not answered. Maybe the nasty notation scared people away, so let me suggest a nicer one:

Let's call the three circles $C_1$, $C_2$, $C_3$, and suppose that $C_i$ is defined by a center point $P_i$, a radius $r_i$, and a unit vector $N_i$ normal to its plane.

So, again, the question is:
find the equations of the tangent planes in terms of the $P_i$, $r_i$, and $N_i$.

The case where the three radii are equal is of some interest, if that's easier.

Also, I'm interested only in the case where the circles are in "general position", which means (I think) that the number of solutions is finite but non-zero. So, please feel free to ignore special cases like the circles having a common tangent line, or being coplanar, or lying on a common cylinder or cone, etc.

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    $\begingroup$ Couldn't it be infinitely many? Imagine 3 circles with a common tangent line. It seems to me that line could be in infinitely many planes. $\endgroup$ – Jens Apr 5 '18 at 13:12
  • $\begingroup$ OK. So let's apply the usual hack -- let's say that the circles are "in general position". I'll add that to the question. $\endgroup$ – bubba Apr 5 '18 at 13:19
  • $\begingroup$ @Jens Yes there can be infinitely many for example the centers are colinear and radii equal. However the principle Conservation of Number says that IF the number is finite then it is constant. Thus, for example through $5$ points there is $1$ conic, there could be infinitely many, but if there is a finite number it is $1$, and never $2$ or $3$ or something. $\endgroup$ – Rene Schipperus Apr 5 '18 at 13:21
  • $\begingroup$ I haven't solved it yet, but the problem can be reduced down to finding a common tangent line to two circles in 3D space. Then the common tangent plane is the plane formed by 3 common tangent lines of each pair of circles. Since there are two tangent lines for each pair (in the most general case), this explains why there are $2^3 = 8$ possible planes. $\endgroup$ – Dylan Apr 5 '18 at 15:53
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    $\begingroup$ @Dylan What do you mean by tangent line to a circle in space, how is this different from just intersecting the circle ? $\endgroup$ – Rene Schipperus Apr 5 '18 at 15:57

Let the circles be $$\vec{a}\cos\theta+\vec{b}\sin\theta+\vec{c}\\ \vec{d}\cos\phi+\vec{e}\sin\phi+\vec{f}\\\ \vec{g}\cos\psi+\vec{h}\sin\psi+\vec{k}$$ Let the unit normal to the plane be $\vec{u}$, where $\vec{u}$ is in the unit sphere. The equation of the plane is $\langle u,x\rangle=p$, where $x$ is the position vector of any point in the plane, and $p$ is a constant.

Parallel planes have the same $u$ ,but different constants. If the plane goes through the origin, then its equation is $(u,x)=0$. Otherwise $(u,x)$ is a different number, but still constant.

Each circle touches the plane at one point, but is otherwise on one side of the plane or the other. So $(u,x)=p$ at that point, but not at other points of the circle. So there is one $\theta$, one $\phi$ and one $\psi$ where the value is $p$. The dot product has the same value at those three points, which gives three equations in the six variables (u counts as two) $\vec{u},\theta,\phi,\psi,p$.
$$\langle u,a\rangle\cos\theta+\langle u,b\rangle\sin\theta+\langle u,c\rangle = p =\\ \langle u,d\rangle\cos\phi+\langle u,e\rangle\sin\phi+\langle u,f\rangle = \\ \langle u,g\rangle\cos\psi+\langle u,h\rangle\sin\psi+\langle u,k\rangle$$

Since the rest of the circle is on one side or other of the plane, either $(u,x) > p$ for all other points of the circle, or $(u,x)<p$ for all other points of the circle. So $(u,x)$ has either a local max or local min, as $\theta$ changes. I differentiated ${\frac d{d\theta}}(u,a\cos\theta+b\sin\theta+c)=0$ and so on, to get three more equations:

$$\langle u,a\rangle\sin\theta=\langle u,b\rangle\cos\theta\\ \langle u,d\rangle\sin\phi=\langle u,e\rangle\cos\phi\\ \langle u,g\rangle\sin\psi=\langle u,h\rangle\cos\psi$$ We can eliminate $\theta,\phi,\psi$ to get $$\sqrt{\langle u,a\rangle^2+\langle u,b\rangle^2}+\langle u,c\rangle = \\ \sqrt{\langle u,d\rangle^2+\langle u,e\rangle^2} + \langle u,f\rangle = \\ \sqrt{\langle u,g\rangle^2+\langle u,h\rangle^2} + \langle u,k\rangle$$ I don't know how to simplify beyond that.

  • $\begingroup$ What do you mean dot of $u$ and a point on the circle ? The dot of $u$ and the vector from the origin to the point of the circle ? $\endgroup$ – Rene Schipperus Apr 5 '18 at 15:27
  • $\begingroup$ Yes. THe equation of the plane is $\langle u,x\rangle=constant$ $\endgroup$ – Michael Apr 5 '18 at 15:48
  • $\begingroup$ I dont understand what you are saying. Do you mean $(u,x)=const$ for ALL points on the circle, or just the point of intersection. Isnt the equation of the plane $(u,x)=0$ as $u$ is the normal ? $\endgroup$ – Rene Schipperus Apr 5 '18 at 15:55
  • $\begingroup$ I get what you are saying now. $\endgroup$ – Rene Schipperus Apr 5 '18 at 16:42
  • $\begingroup$ Well, you have 3 equations in three variables, if you add the equation $|u|=1$. $\endgroup$ – Rene Schipperus Apr 5 '18 at 16:47

Here is a solution from algebraic geometry, (related to ideas from Schubert calculus). The set of all planes in 3-space is a three dimensional projective $\mathbb{P}^3$ space, given by the (projective) coefficients of the equation for a plane
that is $$ax+by+cz+d=0 \leftrightarrow [a,b,c,d]$$ (yes, I know the equation is written affine, it doenst matter).

Now let $C$ denote the variety in $\mathbb{P}^3$ of all planes tangent to a given circle (or more generally conic section, it doesnt change the question). We want to find the number of points in $$C_1\cap C_2\cap C_3$$ where $C_i$ is the same as $C$ with three different circles.

Note that $C_i$ is a surface that is of dimension $2$ so the intersection of three surfaces will be a finite number of points, (just as the intersection of $3$ planes is a point.)

Now Bezout's theorem tells us that the number of such points is the product of the degrees of these surfaces, say $d$. Since the surfaces are all equivalent they all have the same degree. Thus the answer is $d^3$.

It remains to show that $d=2$. For this take the intersection of $C$ with a linear space $L$. The simplest linear space in the space of all planes in a pencil, that is, all planes containing a fixed line. It is clear that if you have fixed circle and all planes through a line there will be two of those planes tangent to the circle. Thus $d=2$ and the answer is $$d^3=2^3=8.$$

  • $\begingroup$ Thanks for confirming my guess about the number of solutions. But actually I want to know the equations of those 8 planes. $\endgroup$ – bubba Apr 5 '18 at 13:36
  • $\begingroup$ You want the equations ? Thats a different question. $\endgroup$ – Rene Schipperus Apr 5 '18 at 13:37
  • $\begingroup$ You need to work of the condition for a plane to be tangent to a circle. $\endgroup$ – Rene Schipperus Apr 5 '18 at 13:38
  • $\begingroup$ > That's a different question. Yes, but it's the question I asked. I modified the question to make this more clear. $\endgroup$ – bubba Apr 5 '18 at 13:42

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