Find the equation of a plane tangent to two spheres

Question

Given the equations of two spheres, how would I find the equation of any plane tangent to the two spheres?

I tried something, but I realized that it failed, and I am not sure where to go from here. I have only basic knowledge of cross product, dot product, etc. and have not yet taken calculus.

My attempt:

I know the centers of the two spheres. I pick any point on the surface of the first sphere. I find the vector from the center of the first sphere to the point I selected. I then scale the center of the second sphere by the vector I just found divided by the radius of the first sphere and multiplied by the radius of the second sphere. Then, I construct the vector from the point I chose on the first sphere to the point I found on the second sphere. I take the cross product of this vector with the vector formed by the centers of the two spheres. I use this as the normal vector for my plane and plug in to get its equation.

I noticed by experimentation that this does not work. Is there a way of solving this problem in a similar manner to what I tried above?

Answer 1

Consider a cross-section of the figure. Then it takes one of the following two forms (here I have called one radius $R$, the other radius $r$, and the distance between the centers $D$)

Case 1: Plane does not pass the between the spheres.

In this case we can calculate $\cos\theta=\dfrac{R-r}{D}$, hence $\theta=\cos^{-1}\dfrac{R-r}{D}$

Case 2: Plane does pass between the spheres.

Similarly, here we can calculate $\sin\theta=\dfrac{R+r}{D}$, hence $\theta=\sin^{-1}\dfrac{R+r}{D}$. However, in this diagram we are not quite so interestd in $\theta$ as we are in $\frac{\pi}{2}-\theta$, the angle of the vector from the center of the first circle perpendicular to the plane. Calling this angle $y$, we have $y=\cos^{-1}\dfrac{R+r}{D}$.

Now we turn to our question. For convenience we will let the center of the first sphere be $(0,0,0)$ and the center of the second sphere as $(D,0,0)$ in $\textit{spherical}$ coordinates ($(x,\theta,\phi)$ meaning distance $x$ from the origin, angle $\theta$ from the $z$-axis, and angle $\phi$ on the projection to the xy-plane (see the first picture here)).

Then for case 1, we can describe a desired plane as the normal plane to the vector $(r,\cos^{-1}\dfrac{R-r}{D},\phi)$ and passing through the point $(r,\cos^{-1}\dfrac{R-r}{D},\phi)$ for $\phi\in[0,2\pi)$.

Similarly for case 2, we can describe a desired plane as the normal plane to the vector $(r,\cos^{-1}\dfrac{R+r}{D},\phi)$ and passing through the point $(r,\cos^{-1}\dfrac{R+r}{D},\phi)$ for $\phi\in[0,2\pi)$.

These are exactly all planes tangent two the two spheres.

Please note for the entirety of this we have assumed $D>R+r$ (or in other words no sphere contains part of the other).