143 If the circle $x^{2}+y^{2}=R^{2}(R>0)$ and the curve $||x|-| y||=1$ have all their common points as the vertices of a regular polygon, then $R$ equals
A. 1
B. 1 or $\sqrt{2}$
C. 1 or $\sqrt{2+\sqrt{2}}$
D. 1 or $2+\sqrt{2}$
143 C. First, draw the graph of $||x|-|y||=1$ in the first quadrant: when $x \geqslant 0, y \geqslant 0$, the graph consists of the rays $x-y=1$ and $y-x=1$. By symmetry about the $x$-axis and $y$-axis, the complete graph can be derived (as shown in Figure 4).
Thus, the regular polygon in question is either a square or an octagon. For the square, it is clear that $R=1$. For the octagon, the side length is $\sqrt{2}$,
$$
R=\sqrt{\left(\frac{\sqrt{2}}{2}\right)^{2}+\left(1+\frac{\sqrt{2}}{2}\right)^{2}}=\sqrt{2+2 \sqrt{2}} .
$$