Triangle of 2 Arcs

Let point $G$ be the midpoint of $DE$ . Then by drawing the radii from the center $O$ to $D$ & $E$ , $\triangle ODE$ will be an isosceles triangle, and then $OG$ will be the bisector of the triangle.

Let the smaller radius = $r$ . Then by Pythagorean theorem, $OD^2 = r^2 = OG^2 + DG^2 = OG^2 + 9$ .

Thus, $OG = \sqrt{r^2 - 9}$ .

Now consider the right $\triangle AOG$ . From the image above, $\sin \theta = \dfrac{OG}{AO} = \dfrac{ \sqrt{r^2 - 9}}{r + 3}$ .

Then by the Center Angle Property , $\angle BOC =2\theta$ .

Hence, by Cosine Rule :

$BC^2 = 11^2 = 121 = OB^2 + OC^2 - 2(OB)(OC)\cos 2\theta$ .

From the trigonometry identity, $\cos 2\theta = 1 - 2\sin^2 \theta = 1- \dfrac{2(r^2 - 9)}{(r + 3)^2}$

$121 = (r+3)^2 + r^2 - 2r(r+3) + \dfrac{4r(r^2 - 9)}{(r + 3)} = 9 + \dfrac{4r(r^2 - 9)}{(r + 3)}$

$121-9 = 112 = \dfrac{4r(r^2 - 9)}{(r + 3)}$

$28(r+3) = r(r^2 - 9) = r^3 -9r$

$0 = r^3 - 37r - 84 = (r-7)(r^2 + 7x + 12) = (r-7)(r+3)(r+4)$

Thus, $r = \boxed{7}$ .