how do you find a point equidistant from three other points?

Awesome question.

Call the three points p1, p2, p3

(in vector form, i.e. p1 = (p1x, p1y, p1z) etc.).

Call c the point equidistant from p1, p2, p3.

c may not always exist, for defective cases:

- (ignore the trivial cases where p1, p2, p3 are not distinct points i.e. you only have 1 or 2 unique points)

- if p1, p2, p3 are collinear on any plane, then c does not exist (radius -> infinity)

- in general a result will exist and is going to be c=s, the center of the sphere of some radius on which all points lie

- it probably makes life a lot easier to first transform the origin to some center (e.g. the centroid (p1+p2+p3)/3, or orthocenter, or some other vector definition of center). Anyway not sure what the transform is, but let's call the resulting new coords p1', p2' , p3'.

Now you know the distances from s to p1', p2', p3' must all be equal (to radius r). This gives three equations in three unknowns (sx, sy, sz). i.e.

|p1'-s| = |p2'-s| = |p3'-s| = r

It may be easier to solve the square of that eqn:

(p1x-sx)^2 + (p1y-sy)^2 + (p1z-sz)^2 =

(p2x-sx)^2 + (p2y-sy)^2 + (p2z-sz)^2 =

(p3x-sx)^2 + (p3y-sy)^2 + (p3z-sz)^2 = r^2

However the 3 eqns seem to be quadratic, not linear. So not sure how to generalize the solution.

Notes:

- in the special 2-dimensional case that p1, p2, p3 lie in a plane AND are all equal distance from some c, then c will be the center of circle containing them. Also c necessarily lies in the same plane.

- in the special 2-dimensional case that p1, p2, p3 lie in a plane BUT not at equal radii of some circle, it will be back to the general sphere solution with c=s and s does NOT lie in the plane. Interesting.

- The final solution is going to be some vector formula, symmetric in p1, p2, p3 containing something symmetric involving cross-products, either:

p1' x p2' x p3'

or else p1' x p2', p2' x p3', p3' x p1'

Ah yes, here it is, determinant equation (28) in

http://mathworld.wolfram.com/Sphere.html

(I did not cheat - i searched for these links afterwards)

Four points uniquely define a sphere. So with only three points you should get an infinite family of spheres.

See Mike Yoder's comment from comp.graphics.algorithms, 2 Dec 1993:

Three points can never determine a unique sphere. If they are collinear and distinct, they do not lie on any circle (and hence not on any sphere either). If they are not collinear, they determine a unique circle, which is contained by an infinity of spheres: think of a ring with balls of varying sizes balanced on it.

...as quoted on this guy's webpage, which also discusses efficient numerical algorithms for sphere estimation:

http://stevehollasch.com/cgindex/geometry/sphere4p...

(A good side question would be: give only three points, you can effectively choose a sphere of any radius you want, greater than some minimum, r>r_min. How do you use this freedom to simply finding ANY solution in the 3 pts case?)

Look for the middle of the three points. :)