Given the circle $x^{2}+y^{2}=1$ and the points $P(1,0),Q(-\lambda,v),R(-1,0)$ with $v=\sqrt{1-\lambda^{2}}$. Given the line $x=t$. On this line, the point $A(t,-\frac{(t-1)v}{\lambda+1})$. Then the point $S(\frac{\lambda(t^{2}+1)+2t}{t^{2}+1+2t\lambda},\frac{v(1-t^{2})}{t^{2}+1+2t\lambda}\ )$. Point $B(t,\frac{(t+1)v}{1-\lambda})$. The circle, diameter $AB$, goes through the two fixed points $(t+\sqrt{t^{2}-1},0)$ and $(t-\sqrt{t^{2}-1},0)$.

A synthetic approach could be this one. It is known that if we let $C := PR \cap QS$ then $\triangle{ABC}$ is autopolar wrt. $\Gamma$. Let $O$ be the center of $\Gamma$. Thus $O$ is the orthocenter of $ABC$ but we also know that line $\ell$ would be the polar of $C$. This already implies that $C$ is a fixed point so basically the problem translates to this: Lemma wrote: Let $\ell, O, C$ be fixed objects (line, point, point). Vary $A, B$ over line $\ell$ such that $O$ is orthocenter of $ABC$. Prove that $(AB)$ passes through two fixed points. It is quite simple, by similarity between $\triangle{ADC}$ and $\triangle{ODB}$ we end up with $AD \cdot BD = CD \cdot OD = \text{fixed}$. This implies that if we create the point $U \in OC$ such that $AU \perp BU$ then we have $DU^2 = AD \cdot BD = \text{fixed}$. Since $D$ is fixed, then $U$ is fixed. Let $V := $ reflection of $U$ over $\ell$. Both $U, V$ are fixed and lying on $(AB)$. We're done!