Example 2 Given in $\triangle A B C$, $A, B, C$ are the three interior angles; $a, b, c$ are the sides opposite to angles $A, B, C$ respectively. It is known that: $2 \sqrt{2}\left(\sin ^{2} A-\sin ^{2} C\right)=(a-b) \sin B$, and the circumradius of $\triangle A B C$ is $\sqrt{2}$.
(1) Find angle $C$;
(2) Find the maximum value of the area $S$ of $\triangle A B C$.
Analysis We first use the Law of Sines to convert the side-angle relationship, and choose an appropriate representation of the area.

Solution (1) $2 \sqrt{2}\left(\sin ^{2} A-\sin ^{2} C\right)=(a-b) \sin B$, and given $2 R=2 \sqrt{2}$, by the Law of Sines we have:
$$
2 \sqrt{2}\left[\left(\frac{a}{2 R}\right)^{2}-\left(\frac{c}{2 R}\right)^{2}\right]=(a-b) \cdot \frac{b}{2 R}
$$

So $a^{2}-c^{2}=a b-b^{2}$, thus $a^{2}+b^{2}-c^{2}=a b$
By the Law of Cosines: $2 a b \cos C=a b$, so $\cos C=\frac{1}{2}, C=\frac{\pi}{3}$
(2) $S=\frac{1}{2} a b \sin C=\frac{\sqrt{3}}{4} a b$, and from (1) we know: $a^{2}+b^{2}-a b=c^{2}=(2 R \sin C)^{2}=6$

So $6=a^{2}+b^{2}-a b \geqslant a b$, thus $a b \leqslant 6$, so $S=\frac{\sqrt{3}}{4} a b \leqslant \frac{3}{2} \sqrt{3}$, equality holds when $a=b$, i.e., when $\triangle A B C$ is an equilateral triangle.