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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!

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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 $$2\int\limits_{0}^{\frac{R\sqrt{3}}{2}}\int\limits_{R-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\,dx\,dy$$

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  • $\begingroup$ I have considered this approach, but I really need to practise my geometry! $\endgroup$ – TreeTreeTree Aug 21 '20 at 16:29
  • $\begingroup$ @TreeTreeTree. Can I help you in some way? $\endgroup$ – zkutch Aug 21 '20 at 16:33
  • $\begingroup$ Yeah, I would really like to know if my value for $\alpha$ is incorrect and if so, why? $\endgroup$ – TreeTreeTree Aug 21 '20 at 16:44
  • $\begingroup$ Can you add link to your geometric drawing? $\endgroup$ – zkutch Aug 21 '20 at 16:49
  • $\begingroup$ Sure! However it was done in paint, so it will not be to scale. $\endgroup$ – TreeTreeTree Aug 21 '20 at 16:56

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