       Re: Sphere formula

• To: mathgroup at smc.vnet.net
• Subject: [mg109460] Re: Sphere formula
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Wed, 28 Apr 2010 01:59:01 -0400 (EDT)

```If you meant 3 points in 2 space (you don't), I think the circle's
equation can be found as follows:

zero = Det@({
{x^2 + y^2, x, y, 1},
{x1^2 + y1^2, x1, y1, 1},
{x2^2 + y2^2, x2, y2, 1},
{x3^2 + y3^2, x3, y3, 1}
})

DeleteCases[
Flatten@CoefficientList[(x - a)^2 + (y - b)^2 - rsq - c zero, {x,
y}], 0] == 0];
Solve[eqns, {a, b, c, rsq}] // Together

(output omitted)

The center is {a, b}, and the radius is Sqrt@rsq.

If you mean 3 points in 3 space (I think you do), you'll need some more
work... but the above might give you ideas.

Bobby

On Tue, 27 Apr 2010 03:05:16 -0500, S. B. Gray <stevebg 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
>

--
DrMajorBob at yahoo.com

```

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