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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))

enter image description here

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)

Quanto
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Aiden Chow
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  • 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. – Aiden Chow Feb 12 '20 at 01:06
  • 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). – L.B. Feb 12 '20 at 02:21
  • Oh, so there's no other way to do it than calculus? :( – Aiden Chow Feb 12 '20 at 02:24
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    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. – Blue Feb 12 '20 at 02:42
  • Is there a proof for the formula shown in the wiki page? Thanks for answering btw. – Aiden Chow Feb 12 '20 at 02:49
  • The formula that says: $A=r^2\text {cos}^{-1}\left(\frac{d^2+r^2-R^2}{2dr}\right)+R^2\text{cos}^{-1}\left(\frac{d^2+R^2-r^2}{2dR}\right)-2\Delta$, where $\Delta = \frac{1}{4}\sqrt{(-d+r+R)(d-r+R)(d+r-R)(d+r+R)}$ – Aiden Chow Feb 12 '20 at 02:52
  • @AidenChow: In a triangle $\triangle ABC$ with sides $a$, $b$, $c$ (with $a$ opposite $A$, etc), the Law of Cosines says that $$c^2=a^2+b^2-2a b\cos C \quad\to\quad C=\cos^{-1}\frac{a^2+b^2-c^2}{2ab}$$ So, the Wikipedia formula simply reads $$r^2\cdot(\angle;\text{between $d$ and $r$})+R^2\cdot(\angle;\text{between $d$ and $R$})-2\cdot(\text{area of};\triangle)$$ The first term gives twice the area of the circular sector subtended by the first angle; likewise, the second term. The last term subtracts-away the triangle area, leaving the area of two circular segments (ie, the lens). – Blue Feb 12 '20 at 03:04

2 Answers2

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enter image description here

Let $CE = h$ and note that $x=AB = AE + EB $, i.e. $$x =\sqrt{r_2^2-h^2} + \sqrt{r_1^2-h^2}$$ Square to get $x^2+r_2^2-r_1^2 = 2x\sqrt{r_2^2-h^2}$; square again to obtain

$$h=\frac1{2x}\sqrt{2x^2r_1^2+2x^2r_2^2+2r_1^2r_2^2-x^4-r_1^4-r_2^4}\tag1$$

Then, the circle sector angles are, respectively

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

and the purple and orange areas are respectively the differences between the corresponding circle sectors and triangles, i.e.

$$S_a = \alpha r_1^2 - h\sqrt{r_1^2-h^2},\>\>\>\>\> S_b = \beta r_2^2 - h\sqrt{r_2^2-h^2}$$

Thus, the area enclosed by the two circles is

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

where $h$ is given above by (1).

Quanto
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2

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

Ataulfo
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  • Unless I misunderstand your solution, you're missing a factor of $1/2$ on sector OPR's area? (in first paragraph) – David P Feb 12 '20 at 02:47
  • @David Peterson.-Yes. I had a lapse because I haven't seen this formula for a long time. Thank you. – Ataulfo Feb 12 '20 at 12:13