If you have the curvatures of 4 circles you can get the curvature of the 5th circle easely by S*V, where
Ci=1/Ri, S is a matrix S=(1 0 0 0; 0 1 0 0; 0 0 0 1; 2 2 -1 2) and V=(C1 C2 C3 C4) initially. So we need the 4th line of S^n that is n(n+1) n(n+1) -n n+1. n=1 => C5
If you have 4 circle curvatures then u can multiply the vector of the curvatures with 4 matrices and generate all the possible circles in an Apollonian gasket. S is actually S3 and corresponds to a new circle tangent to the first, third and fourth circles.