Example 16 Given $x^{2}+y^{2}=25$, find the maximum value of the function $t=\sqrt{18 y-6 x+50}+\sqrt{8 y+6 x+50}$.
In this problem, carefully observe the relationship between the "25" in the given expression and the "50" under the square root in the function: $50=25+25$ and $25=x^{2}+y^{2}$. This inspires us to replace "50" with "25+25" and substitute one "25" with "x^{2}+y^{2}", leading to the following solution.
At this point, from the original function $t=\sqrt{8 y+6 x+50}+\sqrt{8 y-6 x+50}$, we can construct the function
$$
\begin{aligned}
f(x, y) & =\sqrt{x^{2}+y^{2}+8 y+6 x+25}+\sqrt{x^{2}+y^{2}+8 y-6 x+25} \\
& =\sqrt{(x+3)^{2}+(y+4)^{2}}+\sqrt{(x-3)^{2}+(y+4)^{2}},
\end{aligned}
$$

The above expression represents the sum of the distances from a moving point $P(x, y)$ on the circle $x^{2}+y^{2}=25$ to the fixed points $A(-3,-4)$ and $B(3,-4)$ on the circle. To find the maximum value of $t$, it is converted to finding the maximum perimeter of $\triangle A P B$. It is easy to see that the perimeter is maximized when $\triangle A P B$ is an isosceles triangle, at which point the coordinates of $P$ are $(0,5)$, i.e., when $x=0, y=5$, $t_{\max }=6 \sqrt{10}$.