Very nice! Perform an inersion with pole $Y$. The circles $(YRQ),(YPS)$ are turned into lines perpendicular to $AX$, while the circles $(YPQ),(YRS)$ are turned into lines perpendicular to $BC$. This means that $PQRS$ is turned into a parallelogram with sides orthogonal to $AX$ and $BC$, and this parallelogram is cyclic iff its a rectangle, which is equivalent to $AX\perp BC$.

ok let me try . let $PQRS$ be cyclic quadrilateral we prove that $AY$ is altitude. the point $PQ\cap RS$ is on radical axis of circumcirecles of $BSRX (C_1),XQPC (C_2)$(3 circles's radical centre) so now let$QR\cap PS=Z$cause PQRS is cyclic we have $ZR.ZQ=ZS.ZP$so $Z$ is homothecy centre of $C_1,C_2$ so $Z$ is on $BC$. $C_1,C_2$ are tangent at each other at $Y$ so we have $ ZR.ZQ=ZS.ZP=ZY^2$ $ZR.ZQ=ZY^2$ and $ZS.ZP=ZY^2$ is power of $Z$ to circumcircle of $XRYQ$and $ASYP$.we can get from this that $AY $ios perpendicular to $BC$. and other side of problem is easier.and i don't write it.(X is on altitude from A to BC prove that PQRS is cyclic) Actually what a powerfull tool INVERSION is.as grobber shown problem became obvious with Inversion

Actually Iam intested to see your solution Behzad(Med Pa ) I know you are busy these days but If you find a time say how is your solution

it should be on radicalaxis

i took this problem from amir 2 recently.and i tried to solve it by INVERSION .and finally i could solve it.. and im sorprising that grobber solution is as the same as me

amir2 wrote: Actually Iam intested to see your solution Behzad(Med Pa ) I know you are busy these days but If you find a time say how is your solution First thanks amir2. The solution is just angle chasing : We have $PQRS$ is cyclic , so , $\angle QPS$ = $180$ - $\angle SRQ$. So $\angle QPS$ + $\angle YRQ$ = $180$ - $\angle SRY$ = $\angle B$ . but $YRSQ$ is cyclic , so $\angle YRQ$ = $\angle YXQ$ . and note we have $\angle QPS$ = $\angle QCY$ - $\angle SPY$ and we have $ASYP$ is cyclic , so $\angle SPY$ = $\angle SAY$ = $\angle BAY$. So we have $\angle QPS$ + $\angle YRQ$ = $\angle B$ = $\angle YXC$ + $\angle XCY$ - $\angle BAY$ = $\angle BYA$ - $\angle BAY$ = $\angle B$ . so $\angle BYA$ = $90$. So $X$ most be on the altitude of $BC$ .