I tried to construct a circle with radius $x+y$ concentric to the central circle with radius $y$, reducing the question to finding the number of sides of a polygon with the circle of radius $x+y$ being the circumscribed circle, and side length $2x$. But then I realized that the $x$-circles would have to be tangent to their neighboring $x$-circles as well. From there I had essentially no idea how this could be solved.
How many circles with radius $x$ can one fit tangent to a circle with radius $y$
circlesgeometry
Related Solutions
As you have observed, $AB$ and $AC$ are perpendicular, so take $A$ as the origin and assign coordinates $B(4,0)$ and $C(0,3)$.
Let the centre of the inner circle be at $(a,b)$ and the radius is $r$. Then, considering distances, the following equations apply:
$$a^2+b^2=(1+r)^2$$ $$(a-4)^2+b^2=(3+r)^2$$ $$a^2+(b-3)^2=(2+r)^2$$
Subtracting the first equation from the second gives rise to $$a=\frac{2-r}{2}$$ Subtracting the first equation from the third gives rise to $$b=\frac{3-r}{3}$$
Substituting these into the first equation then leads to the quadratic equation $$23r^2+132r-36=0$$
The roots are $$r=\frac{6}{23}$$ which is the answer you are looking for, and $$-6$$ which corresponds to the radius $6$ of the outer kissing circle. All of this can be verified by applying Descartes' Formula.
I've added a picture below. It is trivial to see that the angle between the origins of two circles side by side is $\frac{2\pi}{n}$. Then the angle $\angle{AOB}=\frac{\pi}{n}$. Now we can use the law of sines: $\frac{r}{\sin(\pi/n)}=\frac{|OB|}{\sin(\pi/2)} \implies |OB|=\frac{r}{\sin(\pi/n)}$. Let's call the radius of big circle $R$. Then, $R=|OB|+r = \frac{r}{\sin(\pi/n)} + r$. Then $r=\frac{R}{\frac{1}{\sin(\pi/n)}+1}$. So, we have the radius of small circles. Let polar coordinates of one of the centers be $|R-r|(\sin\theta+\cos\theta)$. Now we can find the coordinates of other centers as well since we know the angle between them is $\frac{2\pi}{n}$. 
Best Answer
\begin{align} \triangle O_0O_1T_0:\quad \sin\tfrac{\pi}n&=\frac{R}{r+R} ,\\ n&=\frac{\pi}{\arcsin(\frac{R}{r+R})} . \end{align}