Geometry circle problem and the curves $y= \pm \frac{1}{x}$ 
A circle touches the $y-$axis at the origin and the curves $y= \frac{1}{x}$,  $y= -\frac{1}{x}$ when $x>0$. Determine the radius of the circle.


I managed to do the question, but it got pretty messy and terrible. By first denoting the center as $C(x_0, y_0)$ we can deduce that the radius is simply $x_0$. Then defining the points $A(a, \frac{1}{a})$ and $B(b, -\frac{1}{b})$ and drawing the tangents from the center to the points where the curves touch the circle I could algebraicly solve this and here it went messy. I had to use the derivative and the property that the product of two gradients equal $-1$ in order to first find point $a$ from what I could use to find $x_0$. Is there some clean way to do this or is it just one of those problems where it gets messy anyway?

It is quite easy. By the symmetry of the problem, we know the circle's center must lie along the $x$-axis. So we only need one coordinate to describe it, which we will call $c$. Then the circle must have equation
$$(x - c)^2 + y^2 = r^2$$
Moreover, since the left hand extremity of the circle should touch the $y$-axis, then a line from the center to that touching point is a radius, so $r = c$. Thus
$$(x - c)^2 + y^2 = c^2$$
and we can expand out to get
$$\begin{align}
(x^2 - 2cx + c^2) + y^2 &= c^2\\
x^2 + y^2 - 2cx + c^2 &= c^2\\
x^2 + y^2 &= 2cx
\end{align}$$
So we now have a growable circle coming off the y-axis, that we can let the little thing grow up (by letting $c$ do so) just as much as need be so he juuust bumps/kisses the curve (❤︎#mehhr). Hence, we must find the limiting case where only a single point of the circle lies upon the curve. So take $y = \pm \frac{1}{x} = \pm x^{-1}$ in the above, to get
$$x^2 + x^{-2} = 2cx$$
Times $x^2$, you get
$$x^4 + 1 = 2cx^3$$
so we want to find which $c$ makes
$$x^4 - 2cx^3 + 1 = 0$$
have only a single solution (coordinate of point of intersection) in $x$. This is most easily done with calculus, though there are purely algebraic methods but a bit nontrivial since this is a quartic (degree 4) polynomial equation. Differentiating, we get that the minimum point is when $4x^3 - 6cx^2 = 0$ so $4x - 6c = 0$ and $x = \frac{6c}{4} = \frac{3c}{2}$. The height of the minimum is then
$$\begin{align}
\left(\frac{3c}{2}\right)^4 - 2c\left(\frac{3c}{2}\right)^3 + 1 &= \frac{81c^4}{16} - 2c\left(\frac{27c^3}{8}\right) + 1\\
 &= \frac{81c^4}{16} - \frac{54c^4}{8} + 1\\
 &= \frac{81c^4}{16} - \frac{108c^4}{16} + \frac{16}{16}\\
 &= \frac{-27c^4 + 16}{16}\end{align}$$
So this height is 0 when $16 = 27c^4$ or $c = r = \sqrt[4]{\frac{16}{27}}$.
Thus the answer is the circle at $\left(\sqrt[4]{\frac{16}{27}}, 0\right)$ and with radius $\sqrt[4]{\frac{16}{27}}$. For scale, this center and radius are approximately 0.877 units, which looks to be about right from your graph given the center is just a bit to the left of 1 on the x-axis. This number seems to be what everybody else is getting, just expressed in different ways.
