Maximizing the side of a square.

$x^2 + y^2 = r^2$ and $x = r - s \implies r^2 - 2sr + s^2 + y^2 = r^2 \implies$

$y = \sqrt{2sr - s^2} \implies R(r - s, \sqrt{2sr - s^2})$ and $m_{OA} = \dfrac{1}{r} \implies y = \dfrac{1}{r}x$

and $W:(r - s, y) \implies y = \dfrac{r - s}{r} \implies W:(r - s, \dfrac{r - s}{r}) \implies$

$s = \overline{WR} = \dfrac{r - s}{r} - \sqrt{2sr - s^2} \implies rs = r - s - \sqrt{2sr - s^2}r \implies$

$(2sr - s^2)r^2 = (r - s(1 + r))^2 = r^2 - 2r(1 + r)s + (1 + r)^2s^2 \implies 2sr^3 - s^2r^2 = r^2 - 2r(1 + r)s + (1 + r)^2s^2 \implies$

$(2r^2 + 2r + 1)s^2 - 2r(r^2 + r + 1)s + r^2 = 0 \implies s = \dfrac{r^2 + r + 1 \pm (r^2 + r)}{2r^2 + 2r + 1}$

$(+) \implies s = r \therefore$ we drop $s = r$ and choose $s = \dfrac{r}{2r^2 + 2r + 1} \implies$

$\dfrac{ds}{dr} = \dfrac{1 - 2r^2}{(2r^2 + 2r + 1)^2} = 0 \implies r = \dfrac{1}{\sqrt{2}}$ and

$-\dfrac{1}{\sqrt{2}} < r < \dfrac{1}{\sqrt{2}} \implies \dfrac{ds}{dr} > 0$ and $r > \dfrac{1}{\sqrt{2}} < 0 \implies$ a max occurs at $r = \dfrac{1}{\sqrt{2}}$

$\implies s_{max} = \dfrac{1}{2 + 2\sqrt{2}} = \dfrac{\sqrt{2} - 1}{2} \approx \boxed{0.2071067811865475 }$ .