# Measure of the central angle for the intersection of two congruent circles of area 18 to have an area of 5

Two congruent circles of area 18 intersect; the region of their intersection has area 5. What is the angle $$\theta$$ of the central angle between the centers of the circles and the points of intersection of the circles?

The radius of the circles is $$\sqrt{18/\pi}$$. $$O$$ is the center of one of the circles, and $$P$$ and $$Q$$ are the points of intersection. $$\mathrm{m}\angle\mathit{POQ} = \theta$$. According to the Law of Cosines, $$\begin{equation*} \left\vert \overline{\mathit{PQ}} \right\vert^{2} = 2\left(\frac{18}{\pi}\right)\Bigl(1 - \cos\theta\Bigr) . \end{equation*}$$ The altitude of $$\triangle\mathit{OPQ}$$ from $$O$$ is $$\begin{equation*} \sqrt{\frac{18}{\pi}} \sin\left(\frac{\pi}{2} - \frac{\theta}{2}\right) = \sqrt{\frac{18}{\pi}} \cos\left(\frac{\theta}{2}\right) . \end{equation*}$$ So, $$\begin{equation*} \frac{\theta}{2\pi}\Bigl(18\Bigr) - \frac{1}{2} \left(2\left(\frac{18}{\pi}\right)\Bigl(1 - \cos\theta\Bigr)\right) \left(\sqrt{\frac{18}{\pi}} \cos\left( \frac{\theta}{2}\right)\right) = \frac{5}{2} . \end{equation*}$$ What is the solution - or an approximation of a solution - to this trigonometric equation?

It easier to do it this way: if radius of circle is $$r$$, the area of sector is $$0.5r^2 \theta$$.

$$A_{\triangle{OPQ}}=0.5r^2 \sin \theta$$.

Area of intersection $$A=r^2 \theta-r^2 \sin \theta=5 \rightarrow \theta-\sin \theta=\frac{5}{r^2}=\frac{5\pi}{18}$$. The solution $$\theta \approx 1.837$$

• Sorry but the area of a circular sector is $r^2 \theta/2$. – Jean Marie May 9 at 17:36
• @JeanMarie: But of course, thanks for correcting me! – Vasya May 9 at 17:41
• Thus we agree finally... – Jean Marie May 9 at 17:42 Fig. 1 : Squares have unit area. We are looking for the value of $$\theta$$ = angle(QOP).

The area in question is the sum of the areas of 2 so-called "lunules", (one of them being delimited by points $$P,I,Q$$) whose formula

$$\dfrac{R^2}{2}(\theta-\sin\theta)$$

Thus, as $$R^2=\frac{18}{\pi}$$, the equation to be solved is :

$$\dfrac{9}{\pi}(\theta-\sin\theta)=\dfrac52$$

or $$\theta-\sin\theta=\dfrac{5\pi}{18}$$

or $$\theta=\underbrace{\sin\theta+\dfrac{5\pi}{18}}_{f(\theta)}$$

Applying fixed point iterative method ($$\theta_{n+1}=f(\theta_n)$$ with $$\theta_0=\pi/2$$), one obtains :

$$\theta_1=1.82744..., \theta_2=1.83991..., \theta_3=1.83667...,$$

converging to $$\Theta= 1.837349240792971...$$ (radians), i.e. approximately 105°.

The convergence factor is the absolute value of $$f'(\Theta)=\cos(\Theta)\approx -0.26$$ ; it means that the absolute value of the error $$e_n=\Theta-\theta_n$$ is divided by approximately four at each step.
Furthermore, as $$f'(\Theta)<0$$, the convergence is alternate.