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A line segment $\overline{AB}$ has a length of $x$. A circle with center $A$ has a radius of $r_1$, and another circle with center $B$ has a radius of $r_2$. Also, $r_1+r_2>x$ and $x,r_1,r_2>0$ and $r_1,r_2<x$. Is it possible to find the area of the region inside both circles? If so, how?

(example graph of problem(Desmos))

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Link to the graph

(I don't know if this is a duplicate or not; I will delete this question if it is a duplicate)

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  • $\begingroup$ I'm not too sure how to approach this problem because I am only in the eighth grade, but I think to connect $A$ and $B$ to the two points where the two circles meet. $\endgroup$ – Aiden Chow Feb 12 at 1:06
  • $\begingroup$ It is definitely possible. If you know calculus, this resumes to calculating an (many) integral(s) (the hard part may be finding the limits of integration). $\endgroup$ – L. B. Feb 12 at 2:21
  • $\begingroup$ Oh, so there's no other way to do it than calculus? :( $\endgroup$ – Aiden Chow Feb 12 at 2:24
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    $\begingroup$ See the formula for the area of an "asymmetric lens" on Wikipedia. No calculus required, but you'll need some basic trig to find the angles involved. $\endgroup$ – Blue Feb 12 at 2:42
  • $\begingroup$ Is there a proof for the formula shown in the wiki page? Thanks for answering btw. $\endgroup$ – Aiden Chow Feb 12 at 2:49
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HINT.-In the attached figure you can calculate the intersection point $P$ and the angles $a$ and $b$. You know the area of a circular sector $OPR$ is given by $\dfrac{r_1^2a}{2}$ where $a$ is in radians of course.

1) Area of triangle $OPO'$minus area of circular sector $O'PS$ = area of sector $OPS$

2)Requested area = 2($\dfrac{r_1^2a}{2}$ minus area of sector $OPS$)

enter image description here

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  • $\begingroup$ Unless I misunderstand your solution, you're missing a factor of $1/2$ on sector OPR's area? (in first paragraph) $\endgroup$ – David P Feb 12 at 2:47
  • $\begingroup$ @David Peterson.-Yes. I had a lapse because I haven't seen this formula for a long time. Thank you. $\endgroup$ – Piquito Feb 12 at 12:13
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Note that

$$AB = AE + EB = \sqrt{AC^2-CE^2} + \sqrt{BC^2-CE^2} $$

Substitute $AB = x$, $BC = r_2$, $AC = r_1$ and $CE = h$ to get $x = \sqrt{r_1^2-h^2} + \sqrt{r_2^2-h^2} $, or

$$x - \sqrt{r_2^2-h^2} = \sqrt{r_1^2-h^2}$$

Square both sides,

$$x^2+r_2^2-r_1^2 = 2x\sqrt{r_2^2-h^2}$$

Square again to obtain $h$,

$$h=\frac1{2x}\sqrt{2x^2r_1^2+2x^2r_2^2+2r_1^2r_2^2-x^4-r_1^4-r_2^4}\tag 1$$

Then, the circle sector angles are,

$$\alpha = \sin^{-1}\frac h{r_1}, \>\>\>\>\>\beta= \sin^{-1}\frac h{r_2}$$

The purple area is the difference between the circle sector of angle $2\alpha$ and the triangle $ACD$, i.e.

$$S_a = \alpha r_1^2 - h\sqrt{r_1^2-h^2}$$

Similarly, the orange area is

$$S_b = \beta r_2^2 - h\sqrt{r_2^2-h^2}$$

Thus, the area inside both circles is

$$S_a+S_b = r_1^2\sin^{-1}\frac h{r_1} + r_2^2 \sin^{-1}\frac h{r_2} - xh$$

where $h$ is given by (1).

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