# Area of intersection between two circles with same radius

I am absolutely sure that there exists loads of other posts about this general type of question. However I could not find one correcting my inevitable mistake.

So I am to find the area of the intersection between two circles with the same radius and the second circles centre on the circumference of the first one.

So I thought that I could make a circle sector by connecting two radii from circle 1 to the points of intersection. This sectors area is $$\frac{r^{2}\alpha}{2}$$ where $$\alpha$$ is the angle, in radians, between the two radii.

Using the law of sines the area of the triangle $$OP_1P_2$$, where $$O$$ is the centre of the first circle, $$P_1$$ and $$P_2$$ are the points of intersection, $$r\times r \times \frac{r}{2}sin(\alpha) = \frac{r^3sin(\alpha)}{2}$$.

Thus the wanted area is $$2(\frac{r^2\alpha}{2} - \frac{r^3sin(\alpha)}{2})$$

Now we need to find $$\alpha$$. We can see a triangle with the sides $$r$$ and $$\frac{r}{2}$$ with the angle $$\frac{\alpha}{2}$$. And again using the law of sines we get $$\alpha = \pm \frac{2\pi}{3}+4k\pi$$. But this is supposedly wrong, from an answer by Alvin Chen link

I am sure that if I have done any trivial errors it must be with this triangle and maybe that it really doesn't have the side $$\frac{r}{2}$$?

Very grateful for a correction of my errors! Thanks.

Edit: A poorly drawn Here's image of the problem!

If you accept decision with double integrals, then let's consider 2 circles $$x^2+y^2=R^2$$ and $$x^2+(y-R)^2=R^2$$. Area you want calculate is $$2\int\limits_{0}^{\frac{R\sqrt{3}}{2}}\int\limits_{R-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\,dx\,dy$$
• Yeah, I would really like to know if my value for $\alpha$ is incorrect and if so, why? – TreeTreeTree Aug 21 '20 at 16:44