The radius of a circle, having minimum area, which touches the curve $$y = 4 - x^{2}$$ and the lines, $$y = |x|$$ is:

Let the radius of the circle with the least area be $$r$$, then, the coordinates of the centre = $$(0, 4 - r)$$.

Since the circle touches the line $$y = x$$ in the first quadrant,

$$\therefore \left| \frac{0 - (4 - r)}{\sqrt{2}} \right| = r \implies r - 4 = \pm r\sqrt{2}$$

$$\implies r = \frac{4}{\sqrt{2} + 1} \text{ or } \frac{4}{1 - \sqrt{2}}$$

But $$r \neq \frac{4}{1 - \sqrt{2}}$$ since $$\left[ \because \frac{4}{1 - \sqrt{2}} < 0 \right]$$

$$\therefore r = \frac{4}{\sqrt{2} + 1} = 4(\sqrt{2} - 1)$$