Now I am supposed to prove the following statement
Given circles $\omega_1$ and $\omega_2$ intersects at $X$ and $Y$. Let $l_1$ be a line through the centre of $\omega_1$ and intersecting $\omega_2$ at points $P$ and $Q$ and let $l_2$ be a line through the centre of $\omega_2$ and intersecting $\omega_1$ at points $R$ and $S$. If $P, Q, R, S$ are concyclic then prove that the centre of the circle passing through these points, passes through the radical axis of $\omega_1$ and $\omega_2$.
Now I have tried using the Ptolemy's theorem and some geometry but how do I relate the centre of the resultant circle to that of the radical axis of given circles?
Suppose line $PQ$ and $RS$ intersect at $T$. The power of $T$ with respect to $ω_1$ is $TR \cdot TS$, and the power of $T$ with respect to $ω_2$ is $TP \cdot TQ$. Because $P, Q, R, S$ are concyclic, then$$ TR \cdot TS = TP \cdot TQ. $$ Thus $T$ is on the radical axis of $ω_1$ and $ω_2$.
Now, because line $PQ$ is the radical axis of $ω_2$ and circle $PQRS$, then $OO_1 ⊥ PQ$, i.e. $OO_1 ⊥ TO_2$. Analogously, $OO_2 ⊥ TO_1$. Thus $O$ is the orthocenter of $△TO_1O_2$, which implies $TO ⊥ O_1O_2$. Since $T$ is on the radical axis of $ω_1$ and $ω_2$, then $O$ is also on the radical axis of $ω_1$ and $ω_2$.