Probably easier than constructing 3D circles, because working mainly on lines and planes:
For each pair of spheres, get the equation of the plane containing their intersection circle, by subtracting the spheres equations (each of the form X^2+Y^2+Z^2+aX+bY+c*Z+d=0). Then you will have three planes P12 P23 P31.
These planes have a common line L, perpendicular to the plane Q by the three centers of the spheres. The two points you are looking for are on this line. The middle of the points is the intersection H between L and Q.
To implement this:
- compute the equations of P12 P23 P32 (difference of sphere equations)
- compute the equation of Q (solve a linear system, or compute a cross product)
- compute the coordinates of point H intersection of these four planes. (solve a linear system)
- get the normal vector U to Q from its equation (normalize a vector)
- compute the distance t between H and a solution X: t^2=R1^2-HC1^2, (C1,R1) are center and radius of the first sphere.
- solutions are H+tU and H-tU

A Cabri 3D construction showing the various planes and line L
answered 2009-09-10T17:12:34.160