What's the measure of the radius of the biggest circle below? For reference: In figure A is the tangency point::
LE=2(TE),
$\overset{\LARGE{\frown}}{AN}=60^o$
$\frac{(TE)^2}{R-r}= 10$

Calculate R.(Answer:80)
My progress

$\triangle AON: \angle AON = 60^o, AO=r, ON=r \implies\\
AN =r \therefore \triangle AON(equilateral)$
$\triangle ATL(right): (LN+r)^2=(3x)^2+(r+OT)^2 \implies\\
LN^2+2LNr+r^2 = 9x^2+r^2+2rOT+OT^2 \rightarrow\\
LN^2+2LNr=9x^2+2rOT+OT^2$
....???
I"freeze" here...
Can I say that A, O. T and O1 are collinear?

First note that $A, O $ and $O_1$ are collinear and as $A, O$ and $T$ are collinear, all four points $A, O, T$ and $O_1$ must be collinear. As $\angle O_1AL = 60^\circ$ and $O_1A = O_1L = R$, $\triangle O_1AL$ is equilateral with side length $R$.
Now there are multiple ways to get to the answer. One approach would be,
$LT = \frac{R \sqrt 3}{2} \implies TE = \frac{R}{2 \sqrt 3}$
$OT = \frac{R}{2} - r$
$OT^2 + TE^2 = OE^2 \implies \left(\frac{R}{2} - r\right)^2 + \left(\frac{R}{2 \sqrt3}\right)^2 = r^2$
And we obtain $R = 3r$
Now we also know that $TE^2 = \frac{R^2}{12} = 10(R-r) = \frac{20 R}{3}$.
$\therefore R = 80$
