## Locus of circumcenters

Given the ellipse (e) with axes a, b (a >b) and equation x²/a² + y²/b² =1 and the real number k >1. Construct the ellipse (f) homothetic to (e) with respect to the origin and with homothety ratio k. From a point A (f) draw the tangents to (e) and define the triangles ABC, ADE whose sides opposite to A are the chords BC, DE of (e) and (f) respectively. Find the locus of the circumcenters P, resp. Q, of triangles ABC and ADE.

The picture shows the two loci. Of particular interest is the second locus (red) of circumcenters of ADE. When the factor k=2 the locus becomes an ellipse. This is used in the discussion about maximal triangles inscribed in an ellipse ( MaximalTrianglesProperties.html ). The following facts are easy to prove:
1) B resp. C are the middles of AE resp. AD.
2) The areas of ABC, ADE are constant and independent from the location of point A on (f).
3) Line OA is the median line of the two triangles.
4) For k=2, ED becomes tangent to (e) and AED is of maximal area inscribed in the ellipse (f).
5) Write the coordinates of points using their eccentric angles: X = (a*cos(u0), b*sin(u0)), A = k*X, B=(a*cos(u1), b*sin(u1)), C = (a*cos(u2), b*sin(u2)). Then u1 = u0+uk, u2 = u0-uk, where cos(uk) = 1/k.

The circumcenter of ABC can be calculated by using the well known formula giving the equation of the circle in terms of the coordinates of the vertices A, B, C.

The coordinates (x0,y0) of the center are related to the coefficients of x and y and given by:

These equations give the parametric description of the locus in terms of the eccentric angle u0 of point A. Relative to property (5) see also the file PolarProperty.html .