How to find a plane that is tangent to 3 spheres?

Question

So there are spheres with radius of 1 centered at (1,2,0), (4,5,0) and (1,3,2). How can one find a plane that is tangent to all 3 spheres? Visually, it looks like as if the spheres are sitting on a plane, but I am not sure how to find the equation of the plane.

Answer 1

The spheres centered at (1,2,0) and (4,5,0) have their tops at $z=1$. The sphere centered at (1,3,2) has its bottom at $z=1$. Therefore, the plane $z=1$ is tangent to all three spheres.

Answer 2

There are actually 8 different solutions, since each sphere can touch the plane from 2 different sides. Two solutions have all 3 spheres touching on one side and are parallel to the plane through the sphere centers.

The other 6 solutions all have two spheres touching on one side and the third sphere touching on the other side. By shifting the plane along its normal vector through the center of the two spheres on the same side the problem can be transformed to finding a plane going through 2 points (sphere centers) and touching a third sphere with unchanged center but radius increased to 2.

To find a plane through 2 points and touching a sphere: take the plane P which is normal to the line L through the 2 points and which contains the center of the sphere. Let this plane P cut the line L in S and the sphere in circle C. The tangents of S to C together with the line L form the two solutions to the transformed fitting problem. The plane for the original problem is gained by shifting the plane back along its normal vector.