Maximum area of an isosceles triangle inscribed in a circle with one vertex fixed at the top of the circle

The circle \(x^2+y^2=25\) has centre \(O\) and radius \(5\). Triangle \(PQR\) has its vertices on this circle at \(P(0,5)\), \(Q(x,y)\) and \(R(-x,y)\). The point \(Q\) can move along the part of the circle in the fourth quadrant, and \(R\) then lies in the third quadrant, as shown in the diagram below.

1. Show that, for the point \(Q\), \(y=-\sqrt{25-x^2}\).[1]
2. Hence, find the area \(A\) of triangle \(PQR\) as an expression in \(x\).[3]
3. Show that \(\frac{\mathrm{d}A}{\mathrm{d}x}=\frac{5\sqrt{25-x^2}+25-2x^2}{\sqrt{25-x^2}}\).[4]
4. Hence or otherwise, find the exact coordinates of \(Q\) when \(A\) is at its maximum.[6]