From: watson@madvax.uwa.oz.au (David Watson) Subject: Re: Sphere from 4 points? Organization: Maths Dept UWA Lines: 23 Distribution: world NNTP-Posting-Host: xanthorrhoea.maths.uwa.edu.au In article <1qkgbuINNs9n@shelley.u.washington.edu>, bolson@carson.u.washington.edu (Edward Bolson) writes: |> Given 4 points (non coplanar), how does one find the sphere, that is, |> center and radius, exactly fitting those points? Finding the circumcenter of a tetrahedron is discussed on page 33 in CONTOURING: A guide to the analysis and display of spatial data, by Dave Watson, Pergamon Press, 1992, ISBN 0 08 040286 0, 321p. Each pair of tetrahedral vertices define a plane which is a perpendicular bisector of the line between that pair. Express each plane in the form Ax + By + Cz = D and solve the set of simultaneous equations from any three of those planes that have a vertex in common (all vertices are used). The solution is the circumcenter. -- Dave Watson Internet: watson@maths.uwa.edu.au Department of Mathematics The University of Western Australia Tel: (61 9) 380 3359 Nedlands, WA 6009 Australia. FAX: (61 9) 380 1028