2. Given that $x, y$ are real numbers, satisfying $x^{2}+y^{2} \leqslant 1$. Then the maximum value of $|1 x^{2}+2 x y-y^{2}|$ is $\qquad$
Notice that, $x^{2}+2 x y-y^{2}=k$ is a hyperbola, with asymptotes
$$
x+(1+\sqrt{2}) y=0, x+(1-\sqrt{2}) y=0 .
$$

Then its conjugate hyperbola is
$$
\begin{array}{l}
y=(\sqrt{2}-1) x, \\
y=-(\sqrt{2}+1) x .
\end{array}
$$

As shown in Figure 12, when the hyperbola is tangent to the circle $x^{2}+y^{2}=1$, $k$ takes the maximum value $\sqrt{2}$.