function [ r, center ] = tetrahedron_circumsphere ( tetra )
%*****************************************************************************80
%
%% tetrahedron_circumsphere() computes the circumsphere of a tetrahedron in 3D.
%
% Discussion:
%
% The circumsphere, or circumscribed sphere, of a tetrahedron is the sphere that
% passes through the four vertices. The circumsphere is not necessarily
% the smallest sphere that contains the tetrahedron.
%
% Surprisingly, the diameter of the sphere can be found by solving
% a 3 by 3 linear system. This is because the vectors P2 - P1,
% P3 - P1 and P4 - P1 are secants of the sphere, and each forms a
% right triangle with the diameter through P1. Hence, the dot product of
% P2 - P1 with that diameter is equal to the square of the length
% of P2 - P1, and similarly for P3 - P1 and P4 - P1. This determines
% the diameter vector originating at P1, and hence the radius and
% center.
%
% Licensing:
%
% This code is distributed under the MIT license.
%
% Modified:
%
% 10 August 2005
%
% Author:
%
% John Burkardt
%
% Reference:
%
% Adrian Bowyer, John Woodwark,
% A Programmer's Geometry,
% Butterworths, 1983.
%
% Input:
%
% real TETRA(3,4) the tetrahedron vertices.
%
% Output:
%
% real R, CENTER(3,1), the center of the
% circumscribed sphere, and its radius. If the linear system is
% singular, then R = -1, CENTER = 0.
%
nrhs = 1;
%
% Set up the linear system.
%
a(1:3,1:3) = ( tetra(1:3,2:4) )';
for j = 1 : 3
a(1:3,j) = a(1:3,j) - tetra(j,1);
end
for i = 1 : 3
a(i,4) = sum ( a(i,1:3).^2 );
end
%
% Solve the linear system.
%
[ a, info ] = r8mat_solve ( 3, nrhs, a );
%
% If the system was singular, return a consolation prize.
%
if ( info ~= 0 )
r = -1.0;
center = zeros ( 3, 1 );
return
end
%
% Compute R, X, Y, Z.
%
r = 0.5 * sqrt ( sum ( a(1:3,3+1).^2 ) );
center(1:3,1) = tetra(1:3,1) + 0.5 * a(1:3,3+1);
return
end