Connect the curve 2

Let the equation of the circle be $x^2+y^2= r^2$ . Where, $r$ is the radius of the circle. Then, we have:

$2x + 2y\dfrac {dy}{dx} = 0\quad \Rightarrow \dfrac{dy}{dx} = - \dfrac{x}{y}$

$f(x)=y=x-\dfrac{1}{x} \quad \Rightarrow \dfrac {dy}{dx} = 1 + \dfrac {1}{x^2}$

At the tangent points,

$-\dfrac {x}{y} = 1 + \dfrac {1}{x^2}\quad \Rightarrow -x = \left( 1 + \dfrac {1}{x^2} \right) y \quad \Rightarrow -x = \left( 1 + \dfrac {1}{x^2} \right) \left( x-\dfrac{1}{x} \right)$

$\Rightarrow -x = x-\dfrac{1}{x} + \dfrac{1}{x} -\dfrac{1}{x^3} \quad \Rightarrow \dfrac {1}{x^3} = 2x \quad \Rightarrow x^4 = \dfrac {1}{2} \quad \Rightarrow x^2 = \dfrac {1}{\sqrt{2}}$

$y=x-\dfrac{1}{x} \quad \Rightarrow y^2=\left( x-\dfrac{1}{x} \right)^2 = x^2-2+\dfrac{1}{x^2} = \dfrac {1}{\sqrt{2}} - 2 + \sqrt{2}$

$r^2 = x^2+y^2 = \dfrac {1}{\sqrt{2}} + \dfrac {1}{\sqrt{2}} - 2 + \sqrt{2} = 2\sqrt{2} - 2$

The area of the circle is given by: $\pi r^2 = \pi (2\sqrt{2} - 2) = 2\pi(-1+\sqrt{2})$ .

$\Rightarrow A+B+C = 2+1+2 = \boxed{5}$

The graph to confirm the answer is as follows:

another way to solve this problem is to find the smallest value of radius so we get

$r^{2}=(x-\frac{1}{x})^{2}+x^{2}$

and condition is $\frac{dr}{dx}=0$

$0=\frac{d}{dx}(2x^{2}+\frac{1}{x^{2}}-2)=4x-\frac{2}{x^{3}}$

then $x=\pm 2^{\frac{1}{4}}$

substitute this value into radius square expression

$r^{2}=2\sqrt{2}-2$

then area is equal to

$2\pi(-1+\sqrt{2})=A\pi(-B+C)$

thus $A+B+C=2+1+2=5$

$f(x)$ has rotational symmetry about the origin and so the circle is centred at the origin.

To find where the circle and the function intersect we have to solve the simultaneous equations

$y=x+\dfrac{1}{x}$

$x^2+y^2=r^2$

Squaring the first of these and substituting into the second gives

$x^2+\left (x-\dfrac{1}{x}\right)^2=r^2$

$\implies x^2+\left (x^2-2+\dfrac{1}{x^2}\right)=r^2$

$\implies 2x^4-(2+r^2)x^2+1=0$

We can interpret this as a quadratic equation in $x^2$ . Because we are looking for points of tangency we want it to have exactly one solution for $x^2$ (which will yield two solutions for $x$ , one positive and one negative). So the discriminant of the equation must be zero, in other words

$\left( 2+r^2\right)^2-8=0$

$\implies r^2=2\sqrt{2}-2=2(-1+\sqrt{2})$

The required area is thus $2\pi (-1+\sqrt{2})$

and so $A+B+C=2+1+2=\boxed{5}$

I guessed that you want A + B + C.

Similar to the last connect the curve posted where the curve was y=x+1/x.The same fundamental applies here as the function being odd is symmetric in opposite quadrant(i.e about origin)