Polar of a point with respect to a circle does not depend on the secants chosen to draw it.

Question

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.

Answer 1

Consider the following figure where the line $p=QG$ is the polar of $P$ respect to the conic obtained with the blue quadrilater $ABCD$. Let $IJEF$ a second quarilater and let $K=p\cap IJ$ and $H=p\cap EF$. Note that the lines $QA$, $QB$, $QP$ and $QG$ form an harmonic group.

Desargues' Involution Theorem. The points of intersection of a line with the three pairs of opposite sides of a complete quadrilateral and a conic section circumscribed about the quadrilateral form four pairs of an involution.

By Desargues' Involution Theorem, applied to the quadrilater $ACDB$ and the line $EF $, there exists one and only one involution such that $E\leftrightarrow F$, $P\leftrightarrow P$ and $QA\cap EF\leftrightarrow QB\cap EF$. Projecting from $Q$, we get a unique involution such that $QE\leftrightarrow QF$, $QA\leftrightarrow QB$ and $QP\leftrightarrow QP$. Since the group $QA$, $QB$, $QP$, $QG$ is harmonic, this involution fixes $QG$. Consequently, the group $QP$, $QH$, $QE$, $QF$ is harmonic and hence $PHEF$ is harmonic. Similarly, by considering the line $IJ $, instead of $EF $, the group $PKIJ$ is harmonic as well.

Since the group $LP$, $LM$, $LE$, $LF$ is harmonic, we must to have $LM=HK=p$, thus proving that $p$ doesn't depends on the choice of the quadrilater.

Answer 2

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, $$I,J,K,L,M,N \quad \text{are collinear}$$