# 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?

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:

$$\begin{split} \lim_{u \rightarrow 1} \, &\frac{2 \left(1 - u\right)\sqrt{1 - u^2}}{\arccos(u) - u \sqrt{1-u^2}} \\ &=\lim_{u \rightarrow 1} \, \frac{2\frac{2u^2 - u - 1}{\sqrt{1 - u^2}}}{-2 \sqrt{1 - u^2}} \\ &=\lim_{u \rightarrow 1} \, -\frac{2u^2 - u - 1}{1 - u^2} \\ &=\lim_{u \rightarrow 1} \, \frac{(2u + 1)(1-u)}{(1 + u)(1-u)} \\ &=\lim_{u \rightarrow 1} \, \frac{(2u + 1)}{(1 + u)} \\ &= \frac{3}{2} \end{split}$$

• 3/2 would be very nice answer for the max value of $B/A$. – Anush Jul 5 '19 at 8:14

To compute the area $$B$$ of the rectangle, we need its width and height.

The width $$w$$ is easy to compute, as we know the widest points are at $$(d-r,0)$$ and $$(r,0)$$, thus $$w=2r-d$$

The height $$h$$ can be computed by noticing that the high/low point lies as $$(d/2,+a)$$ and $$(d/2,-a)$$ width $$a$$ being the corresponding point on the intersection of the two circles.
$$a$$ can be computed by the pythagorean theorem if we look at the triangle from the origin of the left circle $$(0,0)$$, the point $$(d/2,0)$$ and $$(d/2,a)$$. Because the point of interest lies on the circle, the hypothenuse is $$r$$, thus overall we can say $$r^2 = (d/2)^2+a^2$$ and thus $$a = \sqrt{r^2-(d/2)^2}$$

The total height $$h$$ is the distance between $$(d/2,+a)$$ and $$(d/2,-a)$$, that is $$2a$$.

To sum up, the total area of the smallest enclosing rectangle is $$B=wh=2(2r-d)\sqrt{r^2-(d/2)^2}$$

If you look at d in the range $$[0,2r]$$, it is apparent that B will be maximized at d=0. If the same holds for B/A is for someone else to solve! :)

1. If you have two circles of the same radii, then it's easier as the setup is symmetric. They overlap over a region of length $$2d-r$$, so that is how long one side of the rectangle is. This also allows you to find the other side by a basic trigonometric argument which is something you would need to calculate their area of overlap anyway (angle subtended at the centre by the arc etc. etc.). It's not too difficult and you don't necessarily need the exact answer to figure out the second part.

2. If you have two identical circles and distance between their centres is $$0$$, it means the area of overlap is the entire circle so the rectangle is a square. Consider the following exercise:

Take a circle $$C$$ of radius $$r$$. centre $$X$$, and consider a sector $$A$$, where the end points of the arc are $$P$$ and $$Q$$, and the angle subtended at the centre is $$x$$. Let the midpoint of the arc $$PQ$$ be $$R$$. Let $$S=A\backslash PQX$$. Then, can you find the condition on $$x$$ by differentiation or otherwise such that the ratio of $$|PQR|/|S|$$ is the highest?

Note that the area of $$PQR$$ is quite nicely related to your original $$B$$. Can you then relate this to your original problem? I believe it is still going to be a fair bit of tedious calculation but I think this a better way to go about it than plug in actual formulae for $$B$$ and $$A$$; think of it as an easier substitution.