TITLE: Polar of a point with respect to a circle does not depend on the secants chosen to draw it.
QUESTION [6 upvotes]: Consider a circle and a point. To find the polar we draw any two secants and find the diagonal points. Then the polar is the line joining other 2 diagonal points. This is the projective definition of a polar. But how do we know that the polar of the red point is independent of the 2 secants/chords chosen to draw it? Proving this for a circle will be enough to show that its true for a general conic because collinearity and concurrence are preserved in a projective transformation.

REPLY [2 votes]: We want to show that the points $I,J,K,L$ are collinear.

For that, we construct some extra points, $M=(AZ)(CX)$ and $N=(AC)(XZ)$.

Now we see that $A,B,C,X,Y,Z$ are 6 points on a conic and thus we can apply Pascal's theorem!
$$J=(AY)(BX) ,\quad K=(BZ)(CY) \quad \text{and} \quad M=(CX)(AZ) \quad \text{are collinear}$$
Applying it once more,
$$I=(AB)(XY) ,\quad L=(BC)(YZ) \quad \text{and} \quad M=(CX)(AZ) \quad \text{are collinear}$$
We also see that $\bigtriangleup ABC$ and $\bigtriangleup XYZ$ have a center of perspectivity at $P$, so by applying Desargues' Theorem, we get
$$I=(AB)(XY) ,\quad L=(BC)(YZ) \quad \text{and} \quad N=(CA)(ZX) \quad \text{are collinear}$$
But the triangles, $\bigtriangleup AYC$ and $\bigtriangleup XBZ$ also have a center of perspectivity at $P$, and thus we can apply Desargues theorem again.
$$J=(AY)(BX) ,\quad K=(BZ)(CY) \quad \text{and} \quad N=(CA)(ZX) \quad \text{are collinear}$$
Combinig these results, we get,
$$J,K,M,N\quad \text{and}\quad I,L,M,N\quad \text{are collinear}$$
Finally,
$$I,J,K,L,M,N \quad \text{are collinear}$$