Connect the curve

Because of the symmetry of $f(x)$ around the origin, the centre of the circle must lie at $(0,0)$ . Hence the radius is equal to the distance from the origin to $(x,y)=(x,x+1/x)$ , i.e. $\sqrt{x^2+(x+1/x)^2}$ , which is what we have to minimise. Equivalently, we can find the minimum of the expression without the square root, that is:

$x^2+(x+\frac1x)^2=2x^2+2+\frac1{x^2}.$

We set its derivative to zero:

$4x-\frac2{x^3}=0$

$x^4=\frac12$

$x=\pm\frac1{\sqrt[4]{2}}.$

This gives a radius of:

$\sqrt{2x^2+2+\frac1{x^2}} = \sqrt{\frac{2}{\sqrt2}+2+\sqrt{2}} = \sqrt{2(1+\sqrt2)}$

therefore the area of the circle is equal to:

$\pi(2(1+\sqrt2))=2\pi(1+\sqrt2)$

so we obtain $A=2,B=1,C=2$ and hence $A+B+C=\boxed{5}$ .

The simplest way to do it is by using properties of discriminant. Because the circle $x^2+y^2=r^2$ only touches the curves and not crosses it, the discriminant for the equation $x^2 + \left ( x+ \frac 1 x \right )^2 = r^2$ must equal to $0$ .

Expanding and simplifying the equation, we get $x^4 + x^2 (2-r^2) + 1 = 0$ , with $b^2 - 4ac = 0 \Rightarrow (2-r^2)^2 - 4(2) = 0 \Rightarrow r^2 = \sqrt 8 + 2$ . So the area of the circle is simply $\pi r^2 = 2\pi (1 + \sqrt 2)$ . With $A=2,B=1,C=2$ .

Using Lagrange multipliers: