Circles In Circle

Let $R$ be the radius of the outer circle and $r_{S}, r_{L}$ be the radii of the smaller and larger interior circles, respectively. Then $r_{S} + r_{L} = R \Longrightarrow r_{L} = R - r_{S}$ .

Now let $O$ be the origin of the outer circle, $A$ the lower end of the chord and $B$ the midpoint of the chord. Then $|OA| = R, |OB| = R - 2r_{S}$ and $|AB| = 2$ . By Pythagoras we then have that

$|OA|^{2} = |OB|^{2} + |AB|^{2} \Longrightarrow R^{2} = (R - 2r_{S})^{2} + 2^{2}$

$\Longrightarrow R^{2} = R^{2} - 4Rr_{S} + 4r_{S}^{2} + 4 \Longrightarrow Rr_{S} - r_{S}^{2} = 1$ .

Next, the area of the yellow region is

$\pi R^{2} - \pi r_{S}^{2} - \pi r_{L}^{2} = \pi(R^{2} - r_{S}^{2} - (R - r_{S})^{2}) =$

$\pi(R^{2} - r_{S}^{2} - R^{2} + 2Rr_{S} - r_{S}^{2}) = 2\pi(Rr_{S} - r_{S}^{2}) = 2\pi \times 1 = \boxed{2\pi}$ .

maybe oversimplified but if the chord can be anywhere then if it was down the middle of the large circle and the inscribed white circles were identical then the area of the yellow circle is 4 PI and each of the white circles is PI so the yellow area is 4 PI - 2 PI = 2 PI