Linear Algebra and the C Language/a06c

 A homogeneous linear system with as many equations 
 as unknowns has a nontrivial  solution if and only 
 if the determinant  of the matrix is zero.

 Let us calculate the equation of the sphere passing through points P, Q, R and S :   

  c1(x^2 +y^2 +z^2) +c2x +c3y +c4z +c5 = 0

  This same equation with the points 
  P(x1,y1,z1)  Q(x2,y2,z2)  R(x3,y3,z3) and S(x4,y4,z4):
 
  c1(x1^2+y1^2+z1^2)+c2x1+c3y1+c4z1+c5 = 0    
  c1(x2^2+y2^2+z2^2)+c2x2+c3y2+c4z2+c5 = 0   
  c1(x3^2+y3^2+z3^2)+c2x3+c3y3+c4z3+c5 = 0  
  c1(x4^2+y4^2+z4^2)+c2x4+c3y4+c4z4+c5 = 0

 The system of five equations:
 
  c1(x^2 +y^2 +z^2) +c2x +c3y +c4z +c5 = 0
  c1(x1^2+y1^2+z1^2)+c2x1+c3y1+c4z1+c5 = 0    
  c1(x2^2+y2^2+z2^2)+c2x2+c3y2+c4z2+c5 = 0   
  c1(x3^2+y3^2+z3^2)+c2x3+c3y3+c4z3+c5 = 0  
  c1(x4^2+y4^2+z4^2)+c2x4+c3y4+c4z4+c5 = 0 
 
  The determinant of the system:

    |(x ^2+y ^2+z ^2) x  y  z  1|
    |(x1^2+y1^2+z1^2) x1 y1 z1 1|
    |(x2^2+y2^2+z1^2) x2 y2 z2 1| = 0
    |(x3^2+y3^2+z3^2) x3 y3 z3 1|
    |(x4^2+y4^2+z4^2) x4 y4 z4 1|

   The determinant in C language:

    |     1           1  1  1  1|
    |(x1^2+y1^2+z1^2) x1 y1 z1 1|
    |(x2^2+y2^2+z1^2) x2 y2 z2 1| = 0
    |(x3^2+y3^2+z3^2) x3 y3 z3 1|
    |(x4^2+y4^2+z4^2) x4 y4 z4 1|

           c1 (x^2 +y^2 +z^2) +
           c2  x +
           c3  y +
           c4  z +
           c5      = 0
  
  To calculate the coefficients of the equation of the sphere, 
  we use the cofactor expansion along the first row.
 
   cof(R1,C1) (x ^2+y ^2+z ^2)+
   cof(R1,C2)  x+
   cof(R1,C3)  y+
   cof(R1,C4)  z+
   cof(R1,C5)    = 0
   
This equation gives us the equation of the sphere
that passes through the four points  P,  Q,  R and S.