Can we find the areas enclosed between a circle and its tangent?

Can we find the areas enclosed between a circle and its tangent?

Question

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 $- \tan \theta$. Write down the equation of the line $PQR$ and so find the co-ordinates of $P$.

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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 $A(\theta) = B\left(\dfrac{\pi}{2} - \theta\right)$ for any $\theta$. Calculate $A\left(\dfrac{\pi}{4}\right)$.

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1. Show that $A \left(\dfrac{\pi}{3} \right) = \sqrt{3} - \dfrac{\pi}{3}.$

Suggestion

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

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?

Interactive diagram

Download GeoGebra file

The angle $\angle OPQ=\theta$ can be varied using the slider.

The values of $A(\theta)$ and $B(\theta)$ are indicated by the height of the bars on the right.

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

1. 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?

Solution

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

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:

$$\begin{align*} \frac{\cos\theta}{\sin\theta} \times m &= -1, \\ \implies m &= -\frac{\sin\theta}{\cos\theta} \\ &= - \tan\theta, \qquad \text{as required.} \end{align*}$$

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:

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

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

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 $P$:

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

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

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 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 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

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