two tangent circles

Let $AE=x$ , $EO = d$ , and the radius of the large circle $AO=OB=DO=r$ . Then $d = \sqrt{x^2-r^2}$ . Extending $DO$ to intersect the large circle at $P$ . Then by intersecting chord theorem, we have:

$\begin{aligned} DE \cdot EP & = AE \cdot EC \\ r^2 - d^2 & = 8x \\ r^2 - x^2 + r^2 & = 8x \\ \implies r & = \sqrt{\frac {x^2+8x}2} & \implies d = \sqrt{\frac {x^2-8x}2} \end{aligned}$

Now let $FQ$ be perpendicular to $AB$ . We note that $\triangle AEO$ and $\triangle AFQ$ are similar. Then $\dfrac {FQ}{EO} = \dfrac {AF}{AE} \implies FQ = \dfrac {d(x+5)}x$ . Since $FB = \sqrt 2 FQ = \dfrac {\sqrt 2d(x+5)}x$ , and $DF = \sqrt 2r - \dfrac {\sqrt 2d(x+5)}x$ . By intersecting chord theorem again, we have:

$\begin{aligned} DF\cdot FB & = AF \cdot FC \\ \left(\sqrt 2r - \frac {\sqrt 2d(x+5)}x\right)\left(\frac {\sqrt 2d(x+5)}x\right) & = 3(x+5) \\ 2(rdx - d^2(x+5)) & = 3x^2 \\ x^2 \sqrt{x^2-64} - (x^2-8x)(x+5) & = 3x^2 \\ x \sqrt{x^2-64} & = 3x + (x-8)(x+5) \\ & = x^2 - 40 \\ x^2(x^2-64) & = x^4 - 80x^2 + 1600 \\ 16x^2 & = 1600 \\ \implies x & = \boxed {10} \end{aligned}$