Let be a fixed circle, and let be a point outside such that . A variable line through intersects the circle at two points and , such that the quadrilateral is cyclic, where are fixed points on the circle.

Define the following:
- ,
- ,
- is the tangent from to the circle , and is the point of tangency.

**Problem:**
Prove that for all such configurations:
1. The points , , and are collinear.
2. The line is perpendicular to chord .
3. As the line through varies, the point traces a fixed straight line, which is parallel to the isogonal conjugate axis (the so-called *isotropic line*) of the centers and .

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### Outline of a Synthetic Proof:

**1. Harmonic Configuration:**
- Since lie on a circle, their cross-ratio is harmonic:
- The intersection points , and form a well-known harmonic setup along the diagonals of the quadrilateral.

**2. Collinearity of , , :**
- The line is tangent to , and due to harmonicity and projective duality, the polar of passes through .
- Thus, , , and must lie on a common line.

**3. Perpendicularity:**
- Since is tangent at and is a chord, the angle between and chord is right.
- Therefore, line is perpendicular to .

**4. Quasi-directrix of :**
- As the line through varies, the point moves.
- However, all such points lie on a fixed line, which is perpendicular to , and is parallel to the isogonal (or isotropic) line determined by the centers and .

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**Further Questions for Discussion:**
- Can this configuration be extended to other conics, such as ellipses?
- Is there a pure projective geometry interpretation (perhaps using polar reciprocity)?
- What is the locus of point , or of line , as varies?

*This configuration arose from a geometric investigation involving cyclic quadrilaterals and harmonic bundles. Any insights, counterexamples, or improvements are warmly welcomed.*

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f(x)=
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thanks