2023 May TZ2 P1 Q9

A circle with equation \(x^2 + y^2 = 9\) has centre \( (0,0) \) and radius \(3\).
A triangle, \(\mathrm{PQR}\), is inscribed in the circle with its vertices at \(\mathrm{P} \ (-3, 0)\), \(\mathrm{Q} \ (x, y)\) and \(\mathrm{R} \ (x, -y) \), where \(\mathrm{Q}\) and \(\mathrm{R}\) are variable points in the first and fourth quadrants respectively. This is shown in the following diagram.

1. For point \(\mathrm{Q}\), show that \(y = \sqrt{9 - x^2}\).

   [1]

2. Hence, find an expression for \(A\), the area of triangle \(\mathrm{PQR}\), in terms of \(x\).

   [3]

3. Show that \(\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{9 - 3x -2x^2}{\sqrt{9 - x^2}}\).

   [4]

4. Hence or otherwise, find the \(y\)-coordinate of \(\mathrm{R}\) such that \(A\) is a maximum.

   [6]