Correct solutions came from Claudio Baiocchi, Italy; Jerome Cherry, Canada; David Stigant, USA; Philippe Fondanaiche, France; Jack Tseng, USA; Lou Cairoli, USA; Nathan Faber, USA; Farid Lian, Colombia; Bill Webb, USA. An extra round of applause goes to Ron Welch, USA, who researched "fold curves" based at points outside the circle, and based on curves other than circles. He discovered many fascinatingly beautiful curves.

Let C be the center of the circle, P a fixed point in the interior of the circle, Q an arbitrary point on the circumference.  The key observation is that the crease line, in green through points A and B, is the perpendicular bisector of segment PQ.  Let D denote the midpoint intersection and F the intersection of AB and the radius CQ.  The envelope we want is the locus of points traced out by F as Q circumnavigates the circle.  The triangles PDF and QDF are congruent (SAS) so segments PF and QF are equal.  Therefore, 1 = CQ = CF + QF = CF + PF and so the sum of the distances from F to the points C and P is constant.  This is the defining property of an ellipse with foci P and C.