<s>[INST] 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?
 [/INST] 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.
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