Solve for the radius of smaller circle in this image. I adapted this problem from IMO CLASS 10 WORKBOOK this is a very interesting problem. I couldn't solve it please help me in solving this.

I tried to solve it using pythagoras theorem in $\triangle NOO'$ but stuck and could not go further.
 A: Because $\overline{OB}= \overline{OC}$ by definition of $O$ being the circumcenter, $\triangle OBC$ is isosceles, which apex is $\angle BOC= 2\angle BAC = 120^{\circ}$ as it is the central angle corresponding the $\angle BAC$.
By symmetry, $\overline{OM}$ is the height that cuts right in the middle of $\triangle OBC$, bisecting $\angle BOC$. Therefore, $\angle NOO_1 = 60^{\circ}$.
Denote the desired radius of the smaller circle as $r = |NO_1|=|O_1M|$.
Since $\overline{NO_1} \perp \overline{OB}$ by definition of the tangency (circle touches $\overline{OB}$), we know that $\triangle ONO_1$ is also a right triangle with angles $(30^{\circ}, 60^{\circ}, 90^{\circ})$, and the lengths are in the ratio $1:\sqrt3:2$. Namely,
$$\frac{|OO_1|}{|NO_1|} = \frac{\text{hypotenuse}}{\text{long leg}} = \frac2{\sqrt 3} \implies |OO_1| =|NO_1| \frac2{\sqrt 3} = \frac{2r}{\sqrt 3}$$ 
We thus have
$$2=|OM|=|OO_1|+|O_1M| = \frac{2r}{\sqrt 3}+r=r(1 + \frac2{\sqrt 3}) \\
\begin{aligned} 
\implies r = \frac2{\frac2{\sqrt 3}+1} = \frac{2(\frac2{\sqrt 3}-1)}{(\frac2{\sqrt 3}+1)(\frac2{\sqrt 3}-1)} 
&= \frac{\frac4{\sqrt 3} -2}{\frac43 - 1} \quad \scriptsize\text{, multiply by $3$ above and below}\\
&= 4\sqrt 3 - 6\end{aligned}$$
