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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.