Is the following Area of Crescent all right?

Is the following Area of Crescent all right?



In the figure below. There are two overlapping circles and the area of Crescent in Red that I have found is $A_C = fracpi rw2$, where $w$ is the shift from center $'X'$ in blue to $'X'$ in red.



Details:
$$A_C = fracA_elipse - A_circle2 = frac[pi r^2 + pi r w] - pi r^22$$



enter image description here




2 Answers
2



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:
$$A_textcrescent=pi r^2-(2A_textsector EAF-2A_triangle AEF)$$



This is because $A_textsector EAF=A_textsector ECF$, 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 $A=frac12absin C$. Then the area of the crescent can be re-written as:
$$A=pi r^2-alpha r^2+r^2sinalpha\
implies A=r^2(pi-alpha+sin alpha)$$





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 ^_^
– Kashan
Aug 31 at 2:17






This is just Geogebra online.
– John Glenn
Aug 31 at 2:20



Note that for $w=r$ we obtain



$$A_C=fracpi r^22$$



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.





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
– Kashan
Aug 31 at 2:16





You are welcome! Bye
– gimusi
Aug 31 at 7:05



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