Well simple

Found the intersection of the 2 tangent
if none .. then the arc is an half circle of R=1/2 [AB] with center at the
middle

if one intersection . (says I)
for an circular arc exist between A and B then both lengt IA and IB must be
equal.. if not, a curve can exist but not a circular curve ...


To found the center ( says O) you must have
IO perpendicular to AB
and have the following rule





If 2 point of intersection , ...well your are under a sphere ...







web.4707726d501959fba9ce4df50@news.povray.org...
> Is there an easy way to join two points by two circular arcs given their
> tangents?
> Input:
> A,B - endpoints
> a,b - their tangents (oriented parallel to the arcs)
> where A+at and B+bs are skew lines (it's easy to handle the planar case)
> Output:
> C,D - arc centers
> arc normals and radii can be calculated easily from these
> E - intersection point of both points
> together with the arcs' common tangent vector it will be used to cut the
> tori.
>
> I think (after some analysing) it has an infinite number of solutions,
> parametrised by a real number.
>
>

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