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

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.
 A: 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.
A: 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 
$$d^3=2^3=8.$$  
A: I have asked the original question you're referring to, and have abandoned it... until recently (last weekend, as a matter of fact). I think it's solveable, but not so easily. Starting point would be projecting the circle #3 on the plane of circle #2- and getting the common tangent of the circle and resulting ellipse. This problem results in 4th degree polynomial (you can find methods for solving that without numerical methods somewhere around here). After that, there is the problem of the third circle, which I intend to solve by projecting the 'unprojected' tangent to the plane of circle #1 and finding two lines parallel to it that are tangent to that circle (quadratic equation). This should yield 8 solutions.
I could post my progrees when I get to the point it's finished, but I cannot guarantee it will be useful- it will be a mix of spatial vector and analytic planar geometry.
So far, I've worked out the tangent problem, and am preparing to tackle projecting the circle on the plane (should be relatively simple if the centre of projected circle coincides with the ellipse centre- which I assume it will, but will check)
A: This is not a direct answer but describes a converse situation in order to define how the  arbitrary constants play out. Hopefully it  could help model the full scene.
It is assumed that:

*

*(N1,N2,N3) are skew.


*Definition of circle tangency in 3-space... A circle diameter 2a is displaced through radius a in the plane of circle to the tangential position T. Rotation of this displaced circle around axis T generates a torus with T as axis of symmetry, making all circles in any meridional plane tangent to T.
Take three spheres of radius $(R_1,R_2,R_3)$ and place their south poles contacting Horizontal plane HP $z=0$ at arbitrary points $(S_1,S_2,S_3).$ North South polar axis are all parallel to $z-$ axis.
Select three arbitrarily different longitude / meridians  $\phi$ on each sphere. Extend the longitude plane till it intersects HP along red lines $T$.
Rotate each polar longitude circle by an arbitrary amount $\Delta \phi$  around axis $T$ towards HP... out of the sphere.
This has re-created a situation in the question in terms three radii, three arbitrary $\phi$s and three $\Delta \phi$s because the three tilted circles are now skew and tangential to the HP.

