Let $O$ be the center and $r$ be the radius of the circle $\Gamma$. If $PQ \cap RS = \{A\}$ and $PS \cap QR = \{B\}$ where the points $P$, $Q$, $R$, $S$ lie on the circle $\Gamma$; then by Miquel's theorem for the triangle $ABQ$, the circumcircles of the triangles $APS$ and $BRS$ intersect at a point $K$ lying on the side $AB$.

Considering the powers of the points $A$ and $B$ with respect to these circles we obtain $AO^2 - r^2 = AS \cdot AR = AK \cdot AB = AK^2 + AK \cdot KB$ and $BO^2 - r^2 = BS \cdot BP = BK \cdot BA = BK^2 + BK \cdot KA$. In particular, $AO^2 - AK^2 = BO^2 - BK^2$, and $OK$ is perpendicular to $AB$. Then we also have $AK \cdot KB = OK^2 - r^2$. Hence, for any pair $\{A, B\}$ satisfying the conditions of the problem, the circle with diameter $AB$ passes through two points on the line $OK$ which are at a distance $\sqrt{OK^2 - r^2}$ away from the line $\ell$. Since $O$, $K$ and $r$ are independent of $A$ and $B$, these two points are constant and form the intersection set.