In the diagram below is sketched the circle with centre \((1,1)\) and radius \(1\) and a line \(L\). The line \(L\) is tangential to the circle at \(Q\). Further, \(L\) meets the \(y\)-axis at \(R\), and the \(x\)-axis at \(P\) in such a way that the angle \(OPQ\) equals \(\theta\) where \(0<\theta<\dfrac{\pi}{2}\).

Show that the co-ordinates of \(Q\) are

\[(1+ \sin \theta, 1 + \cos \theta),\]

and that the gradient of \(PQR\) is \(- \tan \theta\).

Write down the equation of the line \(PQR\) and so find the co-ordinates of \(P\).

The region bounded by the circle, the \(x\)-axis and \(PQ\) has area \(A(\theta)\); the region bounded by the circle, the \(y\)-axis and \(QR\) has area \(B(\theta)\). (See diagram.)

Explain why

\[A(\theta) = B\left(\dfrac{\pi}{2} - \theta\right)\]

for any \(\theta\).

Calculate \(A\left(\dfrac{\pi}{4}\right)\).

Show that

\[A \left(\dfrac{\pi}{3} \right) = \sqrt{3} - \dfrac{\pi}{3}.\]