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Here is the problem :

We consider two circles that intersect in exactly two points. There $O_1$ the center of the first and $r_1$ its radius. There $O_2$ the center of the second and $r_2$ its radius. We note $d=O_1O_2$. Question: Express the area of ​​the intersection of the two disks with the distances $d$, $r_1$ and $r_2$.

My proof :

I call $ A $ and $ B $ the points of intersection of the two disks and $C$ is the intersection point of $d$ and $AB$ ($O_1C=x$).

I selected a landmark $\mathcal R (O, \vec e_x , \vec e_y )$ such that $O_1$ is centered at $(0,0)$ and $(d, 0)$ for $O_2$

The equation of the two circles are $x^2+y^2=r_1^2$ and $(x-d)^2=y^2=r_2^2$.

We have $x=\frac{d^2+r_1^2-r_2^2}{2d}$, then $$y^2=r_1^2-x^2=r_1^2-\left(\frac{d^2+r_1^2-r_2^2}{2d}\right)^2$$

Writing that $AB=2y$ we have $$AB=\frac{\sqrt{4d^2r_1^2-(d^2-r_2^2+r_1^2)^2}}{d}$$

I am going to calculate the half of the requested area.

Denote $\theta$ the angle of $AO_1B$. The half of the requested area is :

$A_1$=Area (sector)$-$Area (isosceles triangle $O_1$) =$$\frac{\pi r_1^2\theta}{2\pi}-\frac{1}{2}r_1^2\sin(\theta)$$

Furthermore $$\sin\left(\frac{\theta}{2}\right)=\frac{AB}{2r_1}=\frac{\sqrt{4d^2r_1^2-(d^2-r_2^2+r_1^2)^2}}{2dr_1}\qquad \cos\left(\frac{\theta}{2}\right)=\frac{x}{r_1}$$

Therefore $$A_1=r_1^2\arccos\left(\frac{d^2-r_1^2+r_2^2}{2dr_1}\right)-\frac{(d^2-r_1^2+r_2^2)AB}{4d}$$

Using $sin(x)=2\sin(\frac{x}2)\times \cos(\frac{x}2)$.

After some algebra to compute the request area we have $$A=r_1^2\arccos\left(\frac{d^2-r_1^2+r_2^2}{2dr_1}\right)+r_2^2\arccos\left(\frac{d^2-r_2^2+r_1^2}{2dr_2}\right)- \frac{\sqrt{Y}}{2} $$ where $$Y=(-d+r_1+r_2)(d+r_2-r_1)(d-r_2+r_1)(d+r_1+r_2)$$

My Question: Can we proved this result with integrals for exemple ?

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Probably there has been some typo in your formulas. For example, if we set d=3, r1=1, and r2=2, the intersection area is zero. However, your last formula would give $r1^2 \arccos 2+ r2^2 \arccos \frac{1}{2}$, which is clearly wrong. I also have some doubts about the last term $\sqrt{Y}/2$, since calculating the area using integrals yields, apart from the terms containing the arccos, two different radicals that cannot be merged. Could you check your calculations? –  Anatoly Aug 30 at 0:09
I have just posted the result obtained using integrals. I hope that this could facilitate your check. –  Anatoly Aug 30 at 0:12

1 Answer 1

Set the center of the first circle in $(0,0)$ and that of the second circle in $(d,0)$. The equations of the circles are $y=\sqrt{r1^2- x^2}$ and $y=\sqrt{r2^2-(x-d)^2}$. In the segment connecting the two centers, the first and second circle intersect the x-axis in the points $(r1, 0)$ and $(d-r2, 0)$, respectively. From the two equations, we also get that the intesection points between the two circles have an $x$-value of $\displaystyle h=\frac {(d^2+r1^2-r2^2)}{2d}$. Then, to calculate the intersection area $A$, we have to estimate the following integrals:

$$A=\displaystyle 2\int_{d-r2}^{h} \sqrt{r2^2-(x-d)^2} + 2\int_{h}^{r1} \sqrt{r1^2-x^2} $$

The first indefinite integral is $${(x-d)}\sqrt{r2^2-(x-d)^2} + r2^2 \tan^{-1}(\frac {x-d}{\sqrt{r2^2-(x-d)^2}})$$

which calculated between $(d-r2)$ and $h$ gives

$$(h-d)\sqrt{r2^2-(h-d)^2} + r2^2 \tan^{-1}(\frac {h-d}{\sqrt{r2^2-(h- d) ^2}}) +\frac {\pi}{2} r2^2 $$

The second indefinite integral is $$x\sqrt{r1^2-x^2} + r1^2 \tan^{-1} (\frac {x}{\sqrt{r1^2-x^2}})$$

which calculated between $h$ and $r1$ gives

$$-h\sqrt{r1^2-h^2} - r1^2 \tan^{-1}(\frac {h}{\sqrt{r1^2-h^2}})+ \frac {\pi}{2} r1^2$$

Summing the two results we get

$$A=(h-d)\sqrt{r2^2-(h-d)^2} + r2^2 \tan^{-1}(\frac {h-d}{\sqrt{r2^2-(h- d) ^2}}) +\frac {\pi}{2} r2^2 - h\sqrt{r1^2-h^2} - r1^2 \tan^{-1} (\frac {h}{\sqrt{r1^2-h^2}})+ \frac {\pi}{2} r1^2 $$

and substituting $\displaystyle h=\frac {(d^2+r1^2-r2^2)}{2d}$ we obtain

$$A=-\frac {(d^2-r1^2+r2^2)}{2d}\sqrt{r2^2-(\frac {d^2-r1^2+r2^2}{2d}) ^2} - r2^2 \tan^{-1}(\frac {\frac {(d^2-r1^2+r2^2)}{2d}}{\sqrt{r2^2- (\frac {(d^2-r1^2+r2^2)}{2d})^2}}) +\frac {\pi}{2} r2^2 - \frac {(d^2+r1^2-r2^2)}{2d} \sqrt{r1^2-(\frac {d^2+r1^2-r2^2}{2d})^2} - r1^2 \tan^{-1}(\frac {\frac {(d^2+r1^2-r2^2)}{2d}}{\sqrt{r1^2-(\frac {d^2+r1^2-r2^2}{2d})^2}})+ \frac {\pi}{2} r1^2 $$

Now if we set $\frac {d^2-r1^2+r2^2}{2d}=r2 \sin \alpha$, the denominator of the first term containing the arctan becomes equal to $r2 \cos \alpha$. We then can substitute the whole term with $\displaystyle r2^2\arcsin \frac {d^2-r1^2+r2^2}{2dr2}$. Similarly, we can follow the same procedure to substitute the second term containing the arctan with $\displaystyle r1^2\arcsin \frac {d^2+r1^2-r2^2}{2dr1}$.

This leads to

$$A=-\frac {(d^2-r1^2+r2^2)}{2d}\sqrt{r2^2-(\frac {d^2-r1^2+r2^2}{2d}) ^2} - r2^2\arcsin \frac {d^2-r1^2+r2^2}{2dr2} +\frac {\pi}{2} r2^2 - \frac {(d^2+r1^2-r2^2)}{2d} \sqrt{r1^2-(\frac {d^2+r1^2-r2^2}{2d})^2} - r1^2\arcsin \frac {d^2+r1^2-r2^2}{2dr1} + \frac {\pi}{2} r1^2 $$

Further moving the terms containing $\pi/2$ into those containing the $\arcsin$ we finally get

$$A=-\frac {(d^2-r1^2+r2^2)}{2d}\sqrt{r2^2-(\frac {d^2-r1^2+r2^2}{2d}) ^2} + r2^2\arccos \frac {d^2-r1^2+r2^2}{2dr2} - \frac {(d^2+r1^2- r2^2)}{2d} \sqrt{r1^2-(\frac {d^2+r1^2-r2^2}{2d})^2} + r1^2\arccos \frac {d^2+r1^2-r2^2}{2dr1}$$

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