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 \(- \tan \theta\).

Could we add the radius of the circle which meets the tangent to the diagram? What angles do we know now?

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 \(\theta\).

If we draw two diagrams, one with angle \(\theta\) at \(P\), and the other with angle \(\left(\dfrac{\pi}{2} - \theta\right)\) at \(P\), then what do we notice?

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

Could we sketch a new diagram with \(\theta=\dfrac{\pi}{4}\)? Can we calculate the length \(OP\)?

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

Could we sketch a new diagram with \(\theta=\dfrac{\pi}{3}\)?

Maybe \(A \left(\dfrac{\pi}{3} \right)\) could be found by finding the area of two right-angled triangles and subtracting part of the area of the circle?