[asy][asy]defaultpen(fontsize(8)); pair A=(0,0), B=(10,0), X=(0,10), Y=(0,-5), H, P, Q; path S = Circle(midpoint(A--B),abs(A-B)/2); P = intersectionpoints(S, X--B)[1]; Q = intersectionpoints(S, Y--B)[1]; H = extension(P,Q,A,B); draw(S); draw(A--Y--B--X--A--P--Q--A--B); dot(P^^Q^^A^^B^^X^^Y^^H); label("A",A,(-1,0));label("B",B,(1,0));label("P",P,(1,1));label("Q",Q,(1,-1));label("X",X,(-1,0));label("Y",Y,(-1,0));label("H",H,(1,-1));[/asy][/asy]Let $ H=PQ\cap AB$. Claim: As $ X$ and $ Y$ vary on $ L$ such that $ AX\cdot AY = c$, $ H$ remains fixed on $ AB$. It is sufficient to show that $ \frac{[PAQ]}{[PBQ]}$ remains constant. $ \frac{AX}{AB}=\frac{AP}{BP}$, $ \frac{AY}{AB}=\frac{AQ}{BQ}$ $ \implies\frac{c}{AB^2}=\frac{AX\cdot AY}{AB^2}=\frac{AP\cdot AQ}{BP\cdot BQ}=\frac{\frac{1}{2}AP\cdot AQ\sin{PAQ}}{\frac{1}{2}BP\cdot BQ\sin{PBQ}}=\frac{[PAQ]}{[PBQ]}$

Let $ X',\ Y'$ be, the points on the tangent line $ L$ of the given circle $ (O)$ at point $ A,$ as the problem states and also such that $ AX' = AY'$ $ ,(1)$ So, we have that $ (AX)\cdot (AY) = u^{2} = (AX')^{2} = (AY')^{2}$ $ ,(2)$ We denote the points $ P'\equiv (O)\cap BX',\ Q'\equiv (O)\cap BY'$ and let be the point $ H\equiv AB\cap P'Q'.$ We will prove that the line-segment $ PQ,$ always passes through the point $ H.$ From the right angled triangle $ \bigtriangleup ABX',$ we have that $ (BP')\cdot (BX') = (AB)^{2}$ $ ,(3)$ Similarly from the right angled triangle $ \bigtriangleup ABY',$ we have that $ (BQ')\cdot (BY') = (AB)^{2}$ $ ,(4)$ From $ (3),$ $ (4),$ we conclude that the points $ X',\ P',\ Q',\ Y'$ are concyclic, with their circumcircle so be it $ (O'),$ which intersects the line-segment $ AB,$ at points $ M,\ N$ such that $ (BP')\cdot (BX') = (BM)\cdot (BN) = (AB)^{2}$ $ ,(5)$ and $ (AM)\cdot (AN) = u^{2}$ $ ,(6)$ By the same way as before, we can say that $ (BP)\cdot (BX) = (AB)^{2} = (BQ)\cdot (BY)$ $ ,(7)$ from the right angled triangles $ \bigtriangleup ABX,\ \bigtriangleup ABY.$ From $ (5),$ $ (7),$ we conclude that the points $ X,\ P,\ M,\ N$ are concyclic, with circumcircle so be it $ (O_{1}).$ Because of $ (AM)\cdot (AN) = u^{2} = (AX)\cdot (AY)$ $ ,(8)$ we conclude that the point $ Y,$ lies on $ (O_{1}).$ The point $ Q$ now, lies also on the circle $ (O_{1}),$ because of $ (7).$ That is, the circle $ (O_{1}),$ always passes through the constant points $ M,\ N$ and then, we conclude that the line segment $ PQ,$ always passes through the fixed point $ H\equiv AB\cap P'Q',$ as the radical center of the circles $ (O),\ (O'),\ (O_{1})$ and the proof is completed. Kostas Vittas.

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Consider inversion around $A$ with radius $\sqrt{AX\times AY}$ Now, we need to solve the following problem: Quote: Let $A$ be a point on line $l$, and let $B$ be a point such that $AB\perp l$. let $X,Y$ be points such that $AX\times AY=c$ for some positive real constant. If circles $(ABY),(ABX)$ intersect the line through $B$ parallel to $l$ at $P,Q$, show that $(APQ)$ passes through a fixed point. let $Z=(APQ)\cap AB$. We claim $Z$ is the fixed point. Note that $XYPQ$ is a rectangle, so we have that $PB\times BQ=c$. Thus, by power of a point $Z$ is the point such that $AB\times BZ=c$, which is independent of $X,Y$, as desired.