<s>[INST] Ratio of area of intersection of two circles and enclosing rectangle Consider two circles with radius $r$ with distance $d < 2r$ from their centres.  The area of their intersection is given by:
$A = r^2\cos^{-1}\left(\frac{d^2} {2dr}\right) + r^2\cos^{-1}\left(\frac{d^2} {2dr}\right) -
\frac{1}{2}\sqrt{(-d+2r)(d^2)(d+2r)}$
Now consider the smallest rectangle enclosing this intersection and call its area $B$. I have two questions:


*

*What is $B/A$?

*$B/A = 4/\pi$ when $d=0$. Is this its maximum?
 [/INST] The area of a circular sector is given by:
$$A_{\text{circular sector}} = \frac{r^2}{2} (\theta - \sin(\theta))$$
The intersection of two circles is just twice this:
$$A_{\text{intersection}} = r^2 (\theta - \sin(\theta))$$
The height of the rectangle can be found using simple trigonometry:
$$h = 2r \sin\left(\frac{\theta}{2}\right)$$
And the width of the rectangle is twice the height of the circle segment.
$$w = 2 r \left(1 - \cos\left(\frac{\theta}{2}\right)\right)$$
$\theta$ itself can be found also using trigonometry.
$$\theta = 2 \arccos\left(\frac{d}{2r}\right)$$
So the ratio between the area of the intersection and its bounding rectangle is:
$$\begin{split}
R &= \frac{4r^2\left(1 - \cos\left(\frac{\theta}{2}\right)\right)\sin\left(\frac{\theta}{2}\right)}{r^2 \left(\theta - \sin\left(\theta\right)\right)} \\
&= \frac{4 \left(1 - \cos\left(\arccos\left(\frac{d}{2r}\right)\right)\right) \sin\left(\arccos\left(\frac{d}{2r}\right)\right)}{2\arccos\left(\frac{d}{2r}\right) - \sin\left(2\arccos\left(\frac{d}{2r}\right)\right)} \\
\end{split}$$
It is true that $\sin( \arccos(x)) = \sqrt{1 - x^2}$ and $\sin(2 \arccos(x)) = 2x \sqrt{1-x^2}$ so:
$$\begin{split}
R &= \frac{4 \left(1 - \frac{d}{2r}\right)\sqrt{1 - {\left(\frac{d}{2r}\right)}^2}}{2 \left(\arccos\left(\frac{d}{2r}\right) - \frac{d}{2r} \sqrt{1-{\left(\frac{d}{2r}\right)}^2} \right)} \\
&= \frac{2\left(1 - \frac{d}{2r}\right)\sqrt{1 - {\left(\frac{d}{2r}\right)}^2}}{\arccos\left(\frac{d}{2r}\right) - \frac{d}{2r} \sqrt{1-{\left(\frac{d}{2r}\right)}^2}} \\
\end{split}$$
We can take $\frac{d}{2r}$ to be $u$, a dimensionless parameter that describes the relationship between the size of the circles and the distance between them.
$$R = \frac{2 \left(1 - u\right)\sqrt{1 - u^2}}{\arccos(u) - u \sqrt{1-u^2}}$$
This function is monotonically increasing on the interval $u \in [0, 1)$ so $R$ has a maximum in the limit when $u = 1$ (and in particular, the complete overlap case of $u = 0$ where $R = \frac{4}{\pi}$ is actually the minimum).
This value can be found via L'Hopital's rule:
$$\begin{split}
\lim_{u \rightarrow 1} \, &\frac{2 \left(1 - u\right)\sqrt{1 - u^2}}{\arccos(u) - u \sqrt{1-u^2}} \\
&=\lim_{u \rightarrow 1} \,  \frac{2\frac{2u^2 - u - 1}{\sqrt{1 - u^2}}}{-2 \sqrt{1 - u^2}} \\
&=\lim_{u \rightarrow 1} \,  -\frac{2u^2 - u - 1}{1 - u^2} \\
&=\lim_{u \rightarrow 1} \,  \frac{(2u + 1)(1-u)}{(1 + u)(1-u)} \\
&=\lim_{u \rightarrow 1} \,  \frac{(2u + 1)}{(1 + u)} \\
&= \frac{3}{2}
\end{split}$$
</s>