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))
(I don't know if this is a duplicate or not; I will delete this question if it is a duplicate)
On
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
$$h=\frac1{2x}\sqrt{2x^2r_1^2+2x^2r_2^2+2r_1^2r_2^2-x^4-r_1^4-r_2^4}$$
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.
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 $OPS$)