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)

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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

| cite | improve this answer | |
  • $\begingroup$ 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

enter image description here

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).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.