1. Let $x^{2}+y^{2} \leqslant 2$. Then the maximum value of $\left|x^{2}-2 x y-y^{2}\right|$ is $\qquad$ .
$$
\text { Two } 1.2 \sqrt{2}
$$

Let $x=r \cos \theta, \boldsymbol{y}=r \sin \theta, 0 \leqslant \theta<2 \pi$.
From the problem, we know $|r| \leqslant \sqrt{2}$. Therefore,
$$
\begin{array}{l}
\left|x^{2}-2 x y-y^{2}\right| \\
=\left|r^{2} \cos ^{2} \theta-2 r^{2} \sin \theta \cdot \cos \theta-r^{2} \sin ^{2} \theta\right| \\
=\sqrt{2} r^{2}\left|\cos \left(2 \theta+\frac{\pi}{4}\right)\right| \\
\leqslant \sqrt{2} r^{2} \leqslant 2 \sqrt{2}
\end{array}
$$