Example 4. Given: $m x+n y=a, n x-m y=b$, $m^{2}+n^{2}=1$. Find $x^{2}+y^{2}$. (a, b are constants)
Analysis: According to the problem, the desired result is independent of the values of $\mathrm{m}$ and $\mathrm{n}$. If we take a special set of values for $m$ and $n$, we can find the value of $x^{2}+y^{2}$. Naturally, $m=0, n=1$ is a set of values that meets the given conditions. In this case, $y=a, x=b$, so the desired result should be $x^{2}+y^{2}=a^{2}+b^{2}$. This leads to the following simplified solution.

Solution: From the given, we have $(m x+n y)^{2}=a^{2}, (n x-m y)^{2}=b^{2}$. Adding these two equations, we get
$$
(m x+n y)^{2}+(n x-m y)^{2}=a^{2}+b^{2}
$$

which simplifies to $\quad\left(m^{2}+n^{2}\right)\left(x^{2}+y^{2}\right)=a^{2}+b^{2}$.
Since $m^{2}+n^{2}=1$, we have
$$
x^{2}+y^{2}=a^{2}+b^{2}
$$