Example 5 In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $c=10, \frac{\cos A}{\cos B}=\frac{b}{a}=\frac{4}{3}$, and $P$ is a moving point on the incircle of $\triangle A B C$, $d$ is the sum of the squares of the distances from $P$ to the vertices $A, B, C$. Find $d_{\min }+d_{\text {max }}$.
Given, $\frac{\cos A}{\cos B}=\frac{\sin B}{\sin A}, \sin 2 A=\sin 2 B, A+B=\frac{\pi}{2}$, i.e., $C=\frac{\pi}{2}$, and it is easy to know $a=6, b=8$. Establish a Cartesian coordinate system as shown in Figure 7-2, and let $P(x, y)$. Clearly, $(x-2)^{2}+(y-2)^{2}=4$. Therefore,
$$
\begin{aligned}
d & =x^{2}+(y-8)^{2}+(x-6)^{2}+y^{2}+x^{2}+y^{2} \\
& =3 x^{2}+3 y^{2}-12 x-16 y+100 \\
& =3\left[(x-2)^{2}+(y-2)^{2}\right]+76-4 y \\
& =88-4 y=f(y),
\end{aligned}
$$

Since $y \in[0,4]$, by the monotonicity of the linear function, $d_{\text {min }}=f(4)=72, d_{\text {max }}=f(0)=88$, i.e.,
$$
d_{\min }+d_{\max }=160
$$

The radius of the circle $r=\frac{6+8-10}{2}=2$.