<s>[INST] What's the the value of $x$ in the circumference below? For reference:

In the figure; calculate $x$, if $r =\sqrt2$.

(Answer: $x = \sqrt2$)

My progress:
Draw $PO_1\perp HG\:(O_1 \in HG).$
Let $O$ be the center of the largest circle.

Using Euclid's Theorem:
$\triangle OPF:OP^2 = OQ^2+PQ^2-2\cdot OQ\cdot FQ$
$\implies ((R-x)^2 =(R-r_2)^2+(r_2+x)^2-2(R-r_2)((r_2-x)$
$\implies R^2 -2Rx+x^2 = R^2-2Rr_2+r_2^2 +r_2^2+2r_2x+x^2 -2Rr_2+2Rx+2r_2^2-2r_2x$
$\therefore\boxed{
r_2^2-r_2R-Rx = 0}$
$\triangle MJR: ((r_1+r)^2 = IH^2 +(r_1-x)^2$
$\implies r_1^2+2r_1r+r^2=IH^2+r_1^2-2r_1x+x^2$
$\therefore \boxed{2r_1(r+x)-x^2 = IH^2}$
$\triangle PFQ: PQ^2=PF^2+FQ^2 $
$\implies (r_2+x)^2=PF^2 + (r_2-x)^2 $
$\implies r_2^2+2r_2x+x^2=PF^2+r_2^2-2r_2x+x^2$
$\therefore \boxed{4r_2x = PF^2}$
...?
 [/INST] In fact, the two small circles here are called 'Twin circles'. As the name suggests, they are congruent. Here's a proof:

Following the image, $EH\parallel AB$ leads to $FEA,$ $FHB,$ $EDC,$ $ADH$ are straight lines. Also $AF$ is extended to meet $CH$ at $G$.
Since $\angle AFB=\angle ACG=90^\circ$, we can say that $H$ is the orthocenter of $\triangle ABG$. That implies $AI\perp BG$.
Therefore $\angle AIB=\angle ADC=90^\circ\implies BG\parallel CE.$
$\displaystyle \therefore \frac{EH}{AC}=\frac{EG}{AG}=\frac{CB}{AB}\implies EH=\frac{AC\cdot CB}{AB}.$
Similarly, we can prove that the diameter of the other circle is also equal to $\frac{AC\cdot CB}{AB}.$
Thus, you get the answer to your question from here.
</s>