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 \(0<\theta<\dfrac{\pi}{2}\).

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

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

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

We know the centre of the circle is at \((1,1)\).

The situation described, with a right-angled triangle inscribed having one vertex at the centre C of the circle, one vertex at Q and the other vertex directly below Q and directly to the right of C.

From the diagram, we can see that to get from the centre, \(C\), to the point \(Q\) we must travel “\(\sin\theta\)” horizontally and “\(\cos\theta\)” vertically, so

\[ Q= \quad (1 + \sin\theta, 1 + \cos\theta).\]

If we now let the length of \(PQR = \ell\), then \[\begin{align*} \text{gradient of } PQR &= \frac{\text{change in }y}{\text{change in }x} \\ &= \frac{-\ell\sin\theta}{\ell\cos\theta} \\ &= - \tan\theta, \qquad \text{as required.} \end{align*}\]

Alternatively, we could find the gradient of the radius to \(Q\) and (because we know the radius and tangent are perpendicular) we can then find the gradient of the tangent \(PQR\).

The gradient of \(CQ\) is \(\dfrac{\cos\theta}{\sin\theta}\) (the change in \(y\) values \(\div\) the change in \(x\) values). The product of perpendicular gradients is \(-1\).

The tangent \(PQR\) is perpendicular to the radius, therefore: \[\begin{align*} \frac{\cos\theta}{\sin\theta} \times m &= -1, \\ \implies m &= -\frac{\sin\theta}{\cos\theta} \\ &= - \tan\theta, \qquad \text{as required.} \end{align*}\]

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

The line \(PQR\) has gradient \(-\tan\theta\) and goes through the point \((1 + \sin\theta, 1 + \cos\theta)\), and so has equation

\[y - (1 + \cos\theta) = -\tan\theta (x - (1 + \sin\theta)).\]

We’re not asked to give the equation of the line in any particular form, so we can leave it as is.

At the point \(P\) we know \(y=0\). Substituting this value into the equation of the line we find \[\begin{align*} -(1 + \cos\theta) &= -\tan\theta (x - (1 + \sin\theta)) \\ \implies 1 + \cos\theta &= \tan\theta (x - (1 + \sin\theta)) \\ \implies \frac{1 + \cos\theta}{\tan\theta} &= x - (1 + \sin\theta) \\ \implies x &= 1 + \sin\theta + \frac{1 + \cos\theta}{\tan\theta}. \end{align*}\]

We could, if desired, simplify this answer:

\[\begin{align*} x &= 1 + \sin\theta + \frac{1}{\tan\theta} + \frac{\cos\theta}{\tan\theta} \\ &= 1 + \sin\theta + \cot\theta + \frac{\cos^2{\theta}}{\sin\theta} \\ &= 1 + \frac{\sin^2{\theta}}{\sin\theta} + \cot\theta + \frac{\cos^2{\theta}}{\sin\theta} \\ &= 1 + \frac{\left(\sin^2{\theta} + \cos^2{\theta} \right)}{\sin\theta} + \cot\theta \\ &= 1 + \frac{1}{\sin\theta} + \cot\theta \\ &= 1 + \cosec\theta + \cot\theta. \end{align*}\]

If we’d looked at our diagram carefully, we could’ve arrived at the same answer:

Diagram of the triangle OPR with Q and a vertical line down from Q to the x axis drawn. This meets the x axis at a point T. Lengths are marked as follows: OT = 1 + sin theta, QT = 1 + cos theta, TP = (1 + cos theta) over tan theta.

Using \(\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}\), we see \(TP = \dfrac{1 + \cos\theta}{\tan\theta}\), and so

\[x = 1 + \sin\theta + \frac{1 + \cos\theta}{\tan\theta}.\]

Therefore the co-ordinates of \(P\) are \((1 + \cot \theta + \mathrm{cosec \hspace{1mm}} \theta, 0)\).

  1. 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

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

    for any \(\theta\).

Let’s draw two diagrams - one with angle \(\theta\) at \(P\) and the other with angle \(\dfrac{\pi}{2} - \theta\) at \(P\):

Different versions of the original diagram, as described. They are reflections of each other in the line y = x, meaning that the A area in one diagram = the B area in the other.

We can see simply from the symmetry of the diagrams that \(A(\theta) = B\left(\dfrac{\pi}{2} - \theta\right)\).

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

To calculate \(A\left(\dfrac{\pi}{4}\right)\), let’s draw the diagram and see if it might help us.

The original diagram drawn but with theta = pi over 4, so that RP has gradient -1.

By symmetry, we see that \[A\left(\dfrac{\pi}{4}\right) = B\left(\dfrac{\pi}{4}\right).\]

Next, we need to know the lengths \(OP\) and \(OR\) of the triangle \(OPR\).

Using our final answer from (i), we have

\[\begin{align*} OP &= 1 + \frac{1}{\sin \dfrac{\pi}{4}} + \dfrac{\cos \dfrac{\pi}{4}}{\sin \dfrac{\pi}{4}} \\ &= 1 + \sqrt{2} + 1 \\ &= 2 + \sqrt{2}. \end{align*}\]

Our triangle is isosceles, since both the angle at \(P\) and the angle at \(R\) are \(\dfrac{\pi}{4}\), so we have \(OR = OP\) and \[OR = 2 + \sqrt{2} \qquad \text{also.}\]

Now we can calculate the area of the whole triangle \(OPR\).

The area of triangle \(OPR\) is \[\frac{1}{2}(2 + \sqrt{2})^2\] which simplifies to \[\begin{align*} \frac{1}{2} (4& + 4\sqrt{2} + 2) \\ &= \frac{1}{2} (6 + 4\sqrt{2}) \\ &= 3 + 2\sqrt{2}. \end{align*}\]

The shaded area in the diagram below is made up of a square of side length \(1\), and \(\dfrac{3}{4}\) of the circle (of radius \(1\)).

The same as the previous diagram but with the circle and the square with vertices O, C (1,0) and (0,1) shaded; the sides of the square are marked out.

The shaded area is therefore: \[1 \times 1 + \frac{3}{4} \times \pi \times (1)^2 \qquad = \qquad 1 + \frac{3}{4} \pi.\]

The area of \(A\left(\dfrac{\pi}{4}\right)\) can now be found. We have

\[\begin{align*} A\left(\frac{\pi}{4}\right) &= \frac{1}{2} \left(\text{area of triangle }OPR - \text{shaded area} \right) \\ &= \frac{1}{2} \left(3 + 2\sqrt{2} - (1 + \frac{3\pi}{4}) \right) \\ &= \frac{1}{2} \left(2 + 2\sqrt{2} - \frac{3\pi}{4} \right) \\ &= 1 + \sqrt{2} - \frac{3\pi}{8}. \end{align*}\]
  1. Show that

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

The original diagram of the circle and line, with theta = pi over 3, and the line PC and the radii CQ and CS drawn, where S is the point where the circle touches the x axis.

The length \(SP\) can be found from the triangle \(CSP\): we have

\[\begin{align*} \tan {\frac{\pi}{6}} &= \frac{CS}{SP} \\ \frac{1}{\sqrt{3}} &= \frac{1}{SP} \\ \text{so } \quad SP &= \sqrt{3}. \end{align*}\]

Therefore the area of triangle \(CSP\) is \[\frac{1}{2} \times 1 \times \sqrt{3} \quad = \quad \frac{\sqrt{3}}{2}.\]

By symmetry, the area of triangle \(CQP\) is also \(\dfrac{\sqrt{3}}{2}\).

The area of the sector \(CQS\) is \[\frac{1}{3} \times \pi \times (1)^2 \quad = \quad \frac{1}{3} \pi.\]

Therefore we can now find the area \(A\left(\dfrac{\pi}{3}\right)\): we have

\[\begin{align*} A\left(\frac{\pi}{3}\right) &= \text{area of triangle } CSP + \text{area of triangle } CQP - \text{area of sector } CQS \\ &= \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} - \frac{\pi}{3} \\ &= \sqrt{3} - \frac{\pi}{3}, \qquad \text{as required.} \end{align*}\]