A question involving Self Polar Orthogonality

Question

$\textbf{Question:}$ Let ω be a circle and suppose P and Q are points such that P lies on the pole of Q (and hence Q lies on the pole of P). Prove that the circle γ with diameter PQ is orthogonal to ω.

I tried by extending OP to X, where O is the center of ω, such that X is on ω. Then, I tried to write out the expression of power of P with respect to ω, but I am unable to invert or use Pythagoras... $\textbf{Source:}$ Evan Chen, Euclidean Geometry in Mathematical Olympiads.

Answer 1

I will state a lemma and you have to prove that as an exercise.

Suppose $(AB,XY)=-1$. Let $M$ be the midpoint of $AB$. Then $MX \times MY=MA^2=MB^2$.

Now, let $PQ \cap \omega =X,Y$ resp. Thus, $(XY,PQ)=-1$ and define $\odot (PQ) \cap \omega=A,B$ resp. and let $M$ be the midpoint of $PQ$ (hence center of $\odot (PQ)$). By the lemma, $MY×MX=MP^2=MQ^2=MA^2=MB^2$ which implies the orthogonality.

Further Remark :Pythagoras and extending $OP$ after defining the center of $\omega$ is quite irrelevant here, you definitely want to bring a center, but note you also got a circle with a known diameter, and hence considering the center of that circle would be more helpful... the most of the motivation.