Example 12 If $x^{2}+y^{2}=25$, find the maximum value of the function $f(x, y)=\sqrt{8 y-6 x+50}+\sqrt{8 y+6 x+50}$.
$$
\begin{aligned}
f(x, y)= & \sqrt{8 y-6 x+50}+\sqrt{8 y+6 x+50} \\
= & \sqrt{(x-3)^{2}+(y+4)^{2}}+ \\
& \sqrt{(x+3)^{2}+(y+4)^{2}} .
\end{aligned}
$$

Thus, $f(x, y)$ can be seen as the sum of the distances from a point $P(x, y)$ on the circle $x^{2}+y^{2}=25$ to two fixed points $A(-3,-4), B(3,-4)$ on the circle, or as the perimeter of an inscribed triangle with a fixed chord $AB$ as one side, minus the fixed length $AB$, as shown in Figure 5-5. To find the maximum value of $f(x, y)$, we only need to find the maximum perimeter of the inscribed triangle. It is easy to see that the perimeter is maximized when the inscribed triangle is an isosceles triangle $ABC$, with point $C$ having coordinates $(0,5)$. Therefore,
$$
\begin{array}{l}
f(x, y)_{\text {max }}=\sqrt{8 \cdot 5+6 \cdot 0+50}+\sqrt{8 \cdot 5-6 \cdot 0+50} \\
=6 \sqrt{10} .
\end{array}
$$