[alogo] Eye curve

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.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]

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:
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
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 .

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