source: CGBLisp/samples/apollonius2.output@ 1

This output is correct; it says that if u1#0,u2#0,u1^2+u2^2#0 then
there is exactly one solution. In fact, we plug the Basis of second case
into macsyma and get the unique solution for this case:
----------------------------------------------------------------
(C12) solve([ ( - 4 * U1) * X7 + (U1^2), ( - 4 * U2) * X8 + (U2^2), ( - U1^2 - U2^2) * X6 + (U1^2 * U2), (2) * X1 + ( - U1), (2) * X2 + ( - U2), (2) * X3 + ( - U1), (2) * X4 + ( - U2), ( - U1^2 * U2 - U2^3) * X5 + (U1 * U2^3) ],[x1,x2,x3,x4,x5,x6,x7,x8]);

                                                       2            2
             U1       U2       U1       U2        U1 U2           U1  U2
(D12) [[X1 = --, X2 = --, X3 = --, X4 = --, X5 = ---------, X6 = ---------, 
             2        2        2        2          2     2         2     2
                                                 U2  + U1        U2  + U1

                                                                 U1       U2
                                                            X7 = --, X8 = --]]
                                                                 4        4
(C13) 
----------------------------------------------------------------

>(load "apollonius2.lisp")
Loading apollonius2.lisp
Condition:
  Green list: [  ]
  Red list: [  - U1^2 - U2^2, U1, U2 ]
 Basis: [ ( - 4 * U1) * X7 + (U1^2), ( - 4 * U2) * X8 + (U2^2), ( - U1^2 - U2^2) * X6 + (U1^2 * U2), (2) * X1 + ( - U1), (2) * X2 + ( - U2), (2) * X3 + ( - U1), (2) * X4 + ( - U2), ( - U1^2 * U2 - U2^3) * X5 + (U1 * U2^3) ]
Condition:
  Green list: [  - U1^2 - U2^2 ]
  Red list: [ U1, U2 ]
 Basis: [ (U2^3) ]
Finished loading apollonius2.lisp
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