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In the figure below. There are two overlapping circles and the area of Crescent in Red that I have found is $A_{C} = \frac{\pi rw}{2}$, where $w$ is the shift from center $'X'$ in blue to $'X'$ in red.

Details: $$A_C = \frac{A_{elipse} - A_{circle}}{2} = \frac{[\pi r^2 + \pi r w] - \pi r^2}{2}$$

enter image description here

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2 Answers 2

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WLOG, assume both centers lie on the $x$-axis. You can use this diagram afterwards:

enter image description here

Since the area of the circle is $A=\pi r^2$, then the area of the crescent should be: $$A_{\text{crescent}}=\pi r^2-(2A_{\text{sector }EAF}-2A_{\triangle AEF})$$

This is because $A_{\text{sector }EAF}=A_{\text{sector }ECF}$, and so does their corresponding triangle. Since the area of a sector is $A=\frac12r^2 \theta$, with $\theta$ in radians, and the area of the triangle is $A=\frac12ab\sin C$. Then the area of the crescent can be re-written as: $$A=\pi r^2-\alpha r^2+r^2\sin\alpha\\ \implies A=r^2(\pi-\alpha+\sin \alpha)$$

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  • $\begingroup$ hmm...Thanks for your solution. On a sidenote, what software have you used for drawing your figures. It seems very elegant for technical writings. It seems your answer is correct. I will notify it as solution after solving it myself as well ^_^ $\endgroup$
    – SJa
    Aug 31, 2018 at 2:17
  • $\begingroup$ This is just Geogebra online. $\endgroup$
    – John Glenn
    Aug 31, 2018 at 2:20
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Note that for $w=r$ we obtain

$$A_C=\frac{\pi r^2}{2}$$

which seems to be wrong.

Moreover how does the area for the ellipse come in the derivation?

For the general formula refer to Circle overlapping.

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  • $\begingroup$ hmm..thanks for notifying. I thought since both circles are of same radius, a slight shift would mean that they move as if formning ellipse. But I can see how you are right and I am wrong. Thanks $\endgroup$
    – SJa
    Aug 31, 2018 at 2:16
  • $\begingroup$ You are welcome! Bye $\endgroup$
    – user
    Aug 31, 2018 at 7:05

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