> It suffices to know E. From there it is just getting an arc from the
> endpoints and a tangent at one endpoint in a plane (two times of course).
>
> ....
>
> If you are still sure that you have a continuum of solutions, then you can
> eliminate one degree by forcing <E-A,E-A>=<E-B,E-B>. Geometrically this
> means that the point E has the same distance from A and B, which is maybe
> not an esthetically bad way to choose among the possibilities. It adds one
> more equation, this time a polynomial of degree 2. It has the benefit that
> for turning the other equations into polynomials it suffices to multiply
> by <E-A,E-A>, yielding polynomials of degree just 2.
>
> Hope this helps,
>
>   Mark Weyer

Thanks a lot for your answer. Meanwhile I had a similar idea of
parametrising by their common tangent direction (the position can be
calculated as a plane-plane intersection) and comparing the corresponding
tangent points. This yiels one equation for two degrees of freedom, thus
creating the continuum.
Also, I might use your idea of selecting the solution.

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