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


$PG = r_1 - r, PR = r_1 + r$ and so by Pythagoras, it is easy to see that $RG = 2 \sqrt{r r_1}$
Now if $O$ is the center of the circle with radius $R$,
$AG = 2r_1 - r ~$ and
$ |OG| = |AG - AO| = |2r_1 - r - R| ~ $
By Pythagoras, $OG^2 = OR^2 - RG^2$
$\implies (2r_1 - r - R)^2 = (R-r)^2 - 4 r r_1$
Solving, $R r = R r_1 - r_1^2 \implies R r = r_1 (R - r_1)$. But as $R = r_1 + r_2$,
$R r = r_1 r_2 \tag1$
Now $KQ = r_2 - x, QS = r_2 + x$ and we obtain $SK = 2 \sqrt{xr_2}$
$OK = OB - KB = R - (2r_2 - x)$
$OK^2 = (R - 2r_2 + x)^2 = OS^2 - SK^2 = (R-x)^2 - 4 xr_2$
Solving, $Rx = r_2 (R - r_2) = r_1 r_2 \tag2$
From $(1)$ and $(2)$, we conclude that $r = x$.
