Find the radius of the largest circle tangent to x=1, y=0, and y=1/x. Find the largest circle that fits under the graph of $y=\frac{1}{x}$ that is tangent to the x-axis and $x=1$. 
My first approach was the basic approach of trying to find a function for the radius of the circle and maximize it. My issue with this approach is that I was not able to constrain the function enough to maximize it.
I guess am lost on unique characteristics of that tangent point for which I can form a constraint. Here is a graph for visual reference of the circle which I am trying to find. 
tangent circle
Furthermore, if anyone has any input on a general methodology on how to approach problems of this nature (finding circles that are tangent to arbitrary curves), I would very much appreciate it! Thank you
 A: Let $r$ be the radius.

Consider two cases . . .

Case $(1)$:$\;$The circle is centered to the left of the line $x=1$.

Then the center of the circle is the point $P=(1-r,r)$, so the equation of the circle is
$$(x-(1-r))^2+(y-r)^2=r^2$$

Let $T=(t,{\large{\frac{1}{t}}})$ be the point where the circle is tangent to the curve $y={\large{\frac{1}{x}}}$.

Then the line segment $PT$ is perpendicular to the tangent line to the curve $y={\large{\frac{1}{x}}}$, at $x=t$, hence the line $PT$ has slope $t^2$.

Thus, we have the system of equations
$$
(t-(1-r))^2
+\left(
\frac{1}{t}-r
\right)^{\!2}
=r^2
\tag{eq1}
$$
$$
\frac
{{\large{\frac{1}{t}}}-r}
{t-(1-r)}
=
t^2
\tag{eq2}
$$
Solving $(\text{eq}2)$ for $r$ yields
$$r=\frac{1+t^3-t^4}{t(1+t^2)^2}\tag{eq3}$$
Using the above to replace $r$ in $(\text{eq}1)$, and then simplifying, we get
$$2t^5-2t^4+4t^3-4t^2+3t-2=0$$
which has only one real root, $t\approx .8576345859$.

Using the above to replace $t$ in $(\text{eq}3)$, we get $r\approx .7321726551$.

Case $(2)$:$\;$The circle is centered to the right of the line $x=1$.

Then the center of the circle is the point $P=(1+r,r)$, so the equation of the circle is
$$(x-(1+r))^2+(y-r)^2=r^2$$

Let $T=(t,{\large{\frac{1}{t}}})$ be the point where the circle is tangent to the curve $y={\large{\frac{1}{x}}}$.

Then the line segment $PT$ is perpendicular to the tangent line to the curve $y={\large{\frac{1}{x}}}$, at $x=t$, hence the line $PT$ has slope $t^2$.

Thus, we have the system of equations
$$
(t-(1+r))^2
+\left(
\frac{1}{t}-r
\right)^2
=r^2
\tag{eq1$'$}
$$
$$
\frac
{{\large{\frac{1}{t}}}-r}
{t-(1+r)}
=
t^2
\tag{eq2$'$}
$$
Solving $(\text{eq}2')$ for $r$ yields
$$r=\frac{1+t^3-t^4}{t(1-t^2)}\tag{eq3$'$}$$
Using the above to replace $r$ in $(\text{eq}1')$, and then simplifying, we get
$$2t^5-2t^4-4t^3+4t^2+t-2=0$$
which has only one positive real root, $t\approx 1.493386956$.

Using the above to replace $t$ in $(\text{eq}3')$, we get $r\approx .3501328531$.

Thus, there are two possible circles, the largest of which is the one centered to the left of the line $x=1$.
