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 $r = x$.

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.

$AB = 2R\\ r_1+r_2= R\\ AO = R$

Euclid's Th.: $\triangle OO_1O_3$ and $OO_2O_4$

$(R-r)^2=(r_1+r)^2+r_2^2-2(r_1-r)r_2\\R^2-2Rr+r^2 = r_1^2+2r_1r+{r^2}+r_2^2-2r_1r_2+2rr_2\implies\\\boxed{R^2-2Rr = (r_1-r_2)^2+2r(r_1+r_2)}(I)$ $(R-x)^2=(r_2+x)^2+r_1^2-2(r_2-x)r_1=\\ R^2-2Rx+{x^2} =r_2^2+2r_2x+{x^2}+r_1^2-2r_2r_1+2xr_1=\\ \boxed{R^2-2Rx=(r_1-r_2)^2+2x((r_1+r_2)}(II)$

$(I)-(II):2Rx-2Rr=2r(r_1+r_2)+2x(r_1+r_2)=$

$2R(x-r)=(\underbrace{r_1+r_2}_{=R})(2(r+x))$ $\therefore\boxed{\color{red}x-r = r+x \implies x = r}$