Example 7 Given real numbers $x, y$ satisfy $x^{2}+y^{2}=25$. Find the maximum value of the function
$$
f(x, y)=\sqrt{8 y-6 x+50}+\sqrt{8 y+6 x+50}
$$
Solve: Convert 50 in the radical appropriately:
$$
\begin{array}{l}
50=25+25=\left(x^{2}+y^{2}\right)+\left(3^{2}+4^{2}\right). \\
\text { Then } f(x, y) \\
=\sqrt{(x-3)^{2}+(y+4)^{2}}+\sqrt{(x+3)^{2}+(y+4)^{2}} .
\end{array}
$$

Thus, the value of the function $f(x, y)$ can be seen as the sum of the distances from a moving point $P(x, y)$ on the circle $x^{2}+y^{2}=25$ to two fixed points $A(-3,-4)$ and $B(3,-4)$ on the circle, as shown in Figure 3. It is easy to know that when $PA + PB = 2a$ (where $a$ is a constant), the locus of point $P$ is an ellipse with $A$ and $B$ as foci and $a$ as the semi-major axis.

When point $P$ is at the farthest intersection point $C(0,5)$ of the perpendicular bisector of $AB$ and the circumference, the value of $a$ is maximized, and its value is
$$
\begin{array}{c}
f(0,5)=\sqrt{(0-3)^{2}+(5+4)^{2}}+ \\
\sqrt{(0+3)^{2}+(5+4)^{2}} \\
=2 \sqrt{3^{2}+9^{2}}=6 \sqrt{10} .
\end{array}
$$