Re: Sphere formula

*To*: mathgroup at smc.vnet.net*Subject*: [mg109476] Re: Sphere formula*From*: Ray Koopman <koopman at sfu.ca>*Date*: Wed, 28 Apr 2010 07:00:29 -0400 (EDT)*References*: <hr65rj$jq6$1@smc.vnet.net>

On Apr 27, 1:05 am, "S. B. Gray" <stev... at ROADRUNNER.COM> wrote: > 1. The center of a sphere through 4 points has a very nice determinant > form. (http://mathworld.wolfram.com/Sphere.html) What I want is a nice > formula for the center of a sphere through 3 points, where the center is > in the plane of the three points. I have a formula but it's a horrible > mess of hundreds of lines, even after FullSimplify. > > 2. (Unlikely) Is there a way to get Mathematica to put a long formula into a > matrix/determinant form if there is a nice one? > > Any tips will be appreciated. > > Steve Gray If x, Dimensions[x] = {n,m}, contains the coordinates of n points in m dimensions, findcenter[x] will return the coordinates of the center of the best-fitting hypersphere (minimum variance of the distances of the points from the center). findcenter[x_] := Block[{m = Mean@x}, Total[#^2].PseudoInverse[#] & [Transpose@x - m]/2 + m] Here is an example with 3 points in 3 dimensions. r = (* radius * random rotation *) 5 Orthogonalize@RandomReal[NormalDistribution[0,1],{3,3}] {{-4.15687, -2.74134, -0.453282}, {1.72438, -1.9056, -4.28897}, {2.17875, -3.72206, 2.52968}} c = {1, 2, 3}; (* center *) x = (* coordinates of 3 random points with known center & radius *) Table[{Cos@#, Sin@#, 0}& @ RandomReal[{-Pi,Pi}].r + c, {3}] {{0.018473, -1.1359, -0.768653}, {2.51514, 5.25315, 6.48158}, {0.818313, -0.881007, -1.0825}} findcenter[x] (* find the center and the *) Sqrt@Total[(Transpose@x - %)^2] (* distances from the center *) {1., 2., 3.} {5., 5., 5.}