# Area of overlap of two circles

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. Is it possible to find the area of the region inside both circles? If so, how?

(example graph of problem(Desmos))

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

• 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 at 1:06
• 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 at 2:21
• Oh, so there's no other way to do it than calculus? :( – Aiden Chow Feb 12 at 2:24
• 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 at 2:42
• Is there a proof for the formula shown in the wiki page? Thanks for answering btw. – Aiden Chow Feb 12 at 2:49

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 $$OPS$$)

• 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 at 2:47
• @David Peterson.-Yes. I had a lapse because I haven't seen this formula for a long time. Thank you. – Piquito Feb 12 at 12:13

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).