Another tight fit .....

With the radius of the circle being $R$ , by symmetry we know that its center is at $(0,R)$ .

Now let $(a, \frac{1}{a})$ be the circle's point of tangency to the curve $y = \frac{1}{|x|}$ in the first quadrant. Since the tangent line at this point has slope $-\frac{1}{a^{2}}$ , the slope of the normal will be $a^{2}$ . This normal passes through the center of the circle, and thus has the equation $y - R = a^{2}x$ . With $(a, \frac{1}{a})$ being on this line we thus have that

$\frac{1}{a} - R = a^{3} \Longrightarrow R = \frac{1}{a} - a^{3}$ .

Next, since the distance between the center of the circle and $(a, \frac{1}{a})$ is $R$ , we have that

$R = \sqrt{a^{2} + (\frac{1}{a} - R)^{2}} \Longrightarrow R^{2} = a^{2} + a^{6} \Longrightarrow (\frac{1}{a} - a^{3})^{2} = a^{2} + a^{6}$

$\Longrightarrow 1 - 2a^{4} + a^{8} = a^{4} + a^{8} \Longrightarrow a = 3^{-\frac{1}{4}}$ .

Thus $R = \frac{1}{a} - a^{3} = 3^{\frac{1}{4}} - 3^{-\frac{3}{4}} = 0.8773826...$ ,

and so $\lfloor 10000*R \rfloor = \boxed{8773}$ .

Here is a graph of the solution.

Due to symmetry, we can see that the circle touches the x-axis at the origin. The equation of the circle is therefore $(y-R)^2 + x^2 = R^2$ .

Consider the first quadrant of the circle where $x > 0, y> 0$ . Let the angle made by the radius normal to the $y = \dfrac {1}{x}$ with the x-axis as $\theta$ .

Then we have $x = R\cos {\theta}$ and $y = R + R\sin{\theta}$

We note that the gradient of the curve is given by: $\dfrac {dy}{dx} = - \dfrac {1}{x^2}$

And that of the circle is given by: $-\dfrac {1}{\tan{\theta}}$ .

At the point of contact they are equal. $\Rightarrow x^2 = \tan{\theta}$ .

From $x = R\cos {\theta}, \quad y = R + R\sin{\theta} \quad \Rightarrow \dfrac {y}{x} = \dfrac {1+\sin{\theta}} {\cos{\theta}}$

But $\dfrac {y}{x} = \dfrac {1}{x^2} = \dfrac {1}{\tan{\theta}} = \dfrac { \cos {\theta}}{\sin{\theta}}\quad \Rightarrow \dfrac { \cos {\theta}} {\sin{\theta}} = \dfrac {1+ \sin {\theta} } {\cos{\theta}}$

$\Rightarrow \sin{\theta}+ \sin^2 {\theta} = \cos^2{\theta}\quad \Rightarrow \sin{\theta}+ \sin^2 {\theta} = 1 - \sin^2{\theta}$

$\Rightarrow 2\sin^2{\theta}+ \sin{\theta} -1 = 0 \quad \Rightarrow \sin{\theta} = \dfrac {-1 \pm \sqrt(1+8)} {4}$ ,

since $\theta$ is acute $\Rightarrow \sin{\theta} = \frac{1}{2}\quad \Rightarrow \theta = 30^o$ .

Now $x = R\cos{30^o}\quad \Rightarrow x^2 = \frac {3}{4} R^2\quad \Rightarrow \tan{30^o} = \frac {3}{4} R^2 \quad \Rightarrow R^2 = \frac {4}{3\sqrt{3}}$

$\Rightarrow R = \dfrac {2} {3^{\frac{3}{4}}} = 0.877382675$

And the required answer $= \lfloor 0.877382675 \times 10000 \rfloor = \boxed {8773}$