Given the circle c and point A draw secants ABC and take D on AB such that DC/DB = k (constant). The locus of D is the (red) eye curve shown.
From its definition follows that: 1) Every secant throught A intersects the curve at two points {D,F}. 2) The same secant intersects the circle at {B,C} and |DB|=|FC|. 3) The common to DF and BC middle E describes a circle with diameter AO. 4) The ratio ED/EB is constant and equal to m = (k+1)/(k-1). 5) The curve is symmetric on the axis OA.
6) Setting OA = a, R for the radius of the circle the polar equation with center at A is: (r-acos(fi))2 = m2(R2-a2sin2(fi)). 7) In cartesian coordinates centered at A and x-axis AO: (x2+y2)(x2+y2-2ax-m2R2)+a2(x2+m2y2)=0. 8) Inverting this curve with respect to the circle of radius L=|AH|, centered at A we obtain an ellipse. In fact, the inverted radius is r' = L2/r and from (6) we obtain:
The figure displays the ellipse obtained through the inversion with radius L = |AH|. It is characterized by its properties: a) To be tangent to {AH, AI} respectively at points {H, I} and, b) to pass through K', which is the inverse of K, c) its focal points are on the circle through {H,I,A}. See Inversion.html , Ellipse.html .