As before, if you start with 2 intersecting circles (circles I & II, of radii r1 & r2), and then draw 2 more circles (III & IV, with radii r3 & r4) inside the I-II intersection which are tangent to both I & II, then you can construct a closed convex C1 curve of 4 segments. By construction, the line of the _centers_ of I & II is perpendicular to the line of _centers_ of III & IV. Now if you simply start with 4 points (P1, P2, P3, P4), such that the segment P1P2 is perpendicular to the segment P3P4, and the two line segments intersect, then we can construct the configuration of circles as above by choosing appropriate radii. In fact, once we choose 1 of the radii, the others can be calculated. In fact, I think that all of these configurations are equivalent to only 2 configurations, modulo Moebius transforms. The 2 different configurations include one in which circles III & IV intersect, and one in which they do not intersect. --- It would be interesting to find something like the "handles"/"control points" of quadratic & cubic splines to design such curves. If I have a line L1 and a point P1 on that line, and a line L2 and a point P2 on that line, then consider the point of intersection P3 (assume that they intersect). We can construct a "hairpin" P1P3P2, which consists of the 2 line segments P1P3 and P3P2. Now replace P3 with a tiny circular arc, and grow that arc until it touches either P1 or P2 (or both). "Growing the arc" consists of increasing the radius -- but not the angle -- of the arc, so that the ends remain C1 continuous with the P1 and P2 line segments, which also shrink. Our "hairpin" now consists of a line segment and a circular arc, and the join point is C1; the line segment may be only a point. This shows how to construct a spline between 2 given points which must meet those points at a particular angle. We note that if we do this inside of a triangle & minimize the number of segments, we produce the incircle.