1. Given $a \in \mathbf{R}^{+}$, the set $M=|(x, y)| x^{2}+y^{2}=a^{2}|, N=|(x, y)|| x+y \mid$ $+|x-y| \leqslant 2 \mid$, then the necessary and sufficient condition for $M \subseteq N$ is ( ).
A. $0<a \leqslant 1$
B. $0<a \leqslant \sqrt{2}$
C. $1 \leqslant a \leqslant \sqrt{2}$
D. $a \geqslant \sqrt{2}$
1. A

Replace $x$ with $-x$, $y$ with $-y$, the value of $|x+y|+|x-y|$ does not change, which indicates that the figure corresponding to $|x+y|+|x-y| \leqslant 2$ is symmetric about the $x$-axis, $y$-axis, and the origin. Therefore, we only need to draw the graph of $|x+y|+|x-y| \leqslant 2$ in the first quadrant, and the rest can be drawn by symmetry. As shown in Figure 5, it represents a square and its interior with sides parallel to the coordinate axes and side length of 2. Given the condition, the circle represented by $x^{2}-y^{2}=a^{2}$ should be inside the square (including touching the square), so $0<a \leqslant 1$.