Hypercircle

The two curves of $xy=2$ is symmetrical along the line $y=-x$ , with each point on the curve in quadrant I having a corresponding point on the curve in quadrant III equidistant from $y=-x$ . The shortest of these distances will be the radius $r$ of the circle and it is along $y=x$ , along which each curve is symmetrical. The intersections of $xy=2$ and $y = x$ are the points where the circle touches the two curves. The two points are when $xy = x^2 = y^2 = 2$ , or $(\sqrt{2}, \sqrt{2})$ and $(-\sqrt{2}, -\sqrt{2})$ . We note that at these two points the gradients of the tangents to the circle are $-1$ . Let us check that of $xy = 2$ , $\implies \dfrac{dy}{dx} = - \dfrac{2}{x^2}$ , for $x = \pm \sqrt{2}$ , $\dfrac{dy}{dx} = -1$ . Therefore, $r = \sqrt{2+2} = \boxed{2}$ .