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Re: Sphere formula

"S. B. Gray" <stevebg at ROADRUNNER.COM> wrote in message
news:hrbach$hq9$1 at
> On 4/27/2010 1:05 AM, S. B. Gray wrote:
> > 1. The center of a sphere through 4 points has a very nice determinant
> > form. ( 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
> >
> Thanks to everyone who answered my question, but there is a simpler
> answer. I forgot the simple fact that any linear combination of two
> vectors lies in the plane of the two vectors.
> Let the three points be p1,p2,p3. Consider the linear function
> p=b(p2-p1)+c(p3-p1) where b,c are to be determined and p is the desired
> center. Now do
> Solve[{Norm(p-p1)==Norm(p-p2),Norm(p-p1)==Norm(p-p3)},{b,c}].
> This gives b,c and therefore p, which will be equidistant from p1,p2,
> and p3 and lie in their plane. Very simple. (I used (p-p1).(p-p1) etc.
> instead of Norm.)
> Steve Gray

Unless I misinterpreted something in this posting, the following does not
give the expected center point {1, 2, 3} using Ray Koopman' s posted values
for {p1, p2, p3} as the result:




Another test using a random center point and Ray Koopman's method of
selecting three random points also does not match the expected value:

  {pc, #, pc == #} &[With[{p=b(p2-p1)+c(p3-p1)},



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