Let $ O$ be the center of $ \Gamma$ and let the diagonals $ PR, SQ$ of $ PQRS$ intersect at $ C.$ $ l \equiv AB$ is a polar of $ C$ WRT the circle $ \Gamma$ $ \Longrightarrow$ $ C$ is fixed. Likewise, $ BC, CA$ are polars of $ A, B.$ $ AO \perp BC, BO \perp CA, CO \perp AB$ $ \Longrightarrow$ $ O$ is orthocenter of $ \triangle ABC.$ Let $ A' \in BC, B' \in CA, C' \in AB$ be feet of the altitudes $ AO, BO, CO.$ Inversion in $ \Gamma$ takes $ A, B$ into $ A', B'.$ The angles $ \angle AA'B, \angle AB'B$ are right, $ A', B'$ are on a circle $ \Delta$ with diameter $ AB.$ Therefore, inversion in $ \Gamma$ takes the circle $ \Delta$ into itself, which means that $ \Delta$ is perpendicular to the inversion circle. The circle $ \Delta$ centered on the line $ \l$ is also perpendicular to this line. The non-intersecting line $ \l$ and circle $ \Gamma$ can be transformed by some other inversion into 2 concentric circles $ \Lambda^*, \Gamma^*.$ Image of $ \Delta$ in this inversion is a circle or line perpendicular to these 2 concentric circles, therefore it is a line $ d^*$ through their common center $ X^*.$ Set of all circles $ \Delta$ goes into set of all lines $ d^*$ through $ X^*,$ therefore, set of all circles $ \Delta$ is a pencil circles of intersecting at $ X, Y$ (and centered on the line $ l$). $ XY \perp l$ and by symmetry, $ O \in XY.$

sinankaral53 wrote: A circle $ \Gamma$ and a line $ \ell$ is given in a plane such that $ \ell$ doesn't cut $ \Gamma$.Determine the intersection set of the circles has $ [AB]$ as diameter for all pairs of $ \left\{A,B\right\}$ (lie on $ \ell$) and satisfy $ P,Q,R,S \in \Gamma$ such that $ PQ \cap RS = \left\{A\right\}$ and $ PS \cap QR = \left\{B\right\}$ turkish solution

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