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 $
$\implies r_2^2+2r_2x+x^2=PF^2+r_2^2-2r_2x+x^2$
$\therefore \boxed{4r_2x = PF^2}$
...?


$AB = 2R\\
r_1+r_2= R\\
AO = R$
Euclid's Th.: $\triangle OO_1O_3$ and $OO_2O_4$
$(R-r)^2=(r_1+r)^2+r_2^2-2(r_1-r)r_2\\R^2-2Rr+r^2 = r_1^2+2r_1r+{r^2}+r_2^2-2r_1r_2+2rr_2\implies\\\boxed{R^2-2Rr = (r_1-r_2)^2+2r(r_1+r_2)}(I)$
$(R-x)^2=(r_2+x)^2+r_1^2-2(r_2-x)r_1=\\
R^2-2Rx+{x^2} =r_2^2+2r_2x+{x^2}+r_1^2-2r_2r_1+2xr_1=\\
\boxed{R^2-2Rx=(r_1-r_2)^2+2x((r_1+r_2)}(II)$
$(I)-(II):2Rx-2Rr=2r(r_1+r_2)+2x(r_1+r_2)=$
$2R(x-r)=(\underbrace{r_1+r_2}_{=R})(2(r+x))$
$\therefore\boxed{\color{red}x-r = r+x \implies x = r}$
