An isosceles triangle \(ABC\) with \(AB = AC\) has \(AD\) as its altitude where \(D\) is on \(BC\). Let \(C^{"}\) be the diametrically opposite point to \(C\) on the circumcircle of \(\Delta ABC\) and let \(M\) be the midpoint of \(AB\).

Construct a circle centred at \(M\) of radius \(MC^{"}\) and let it intersect \(AB\) *(possibly extended)* at \(P\) and \(Q\). Let \(S\) and \(R\) be respectively the reflections of \(P\) and \(Q\) across altitude \(AD\).

Being an isosceles trapezium, quadrilateral \(PQRS\) is cyclic of course. Find the radius \(\rho\) of the circle circumscribing it.

Given: The altitude \(AD = 9\) and circumradius of \(\Delta ABC\) is \(8\).

Report \(\rho^2\).

\(Note:\) An original twist to the famous problem! And by the way, the actual diagram looks not at all like the one shown here for guidance: Figure not to scale in the least!

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