15. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with left and right foci $F_{1}, F_{2}$, and the right vertex $A$, and $P$ is any point on the ellipse $C$. It is known that the maximum value of $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ is 3, and the minimum value is 2.
(1) Find the equation of the ellipse $C$;
(2) If the line $l: y=k x+m$ intersects the ellipse $C$ at points $M, N$ (where $M, N$ are not the left or right vertices), and the circle with diameter $M N$ passes through point $A$. Prove that the line $l$ passes through a fixed point, and find the coordinates of this fixed point.
(1) Let $P(a \cos \theta, b \sin \theta), F_{1}(-c, 0), F_{2}(c, 0)$, then $\overrightarrow{P F_{1}}=(-c-a \cos \theta,-b \sin \theta)$,
$\overrightarrow{P F_{2}}=(c-a \cos \theta,-b \sin \theta)$, thus $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=a^{2} \cos ^{2} \theta-c^{2}+b^{2} \sin ^{2} \theta=c^{2} \cos ^{2} \theta+b^{2}-c^{2}$, hence $\left\{\begin{array}{l}a^{2}-c^{2}=3, \\ b^{2}-c^{2}=2\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=4, \\ b^{2}=3 .\end{array}\right.\right.$ Therefore, the equation of the ellipse $C$ is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right), A(2,0)$, solving the system of equations gives
$$
\left\{\begin{array}{l}
y=k x+m, \\
3 x^{2}+4 y^{2}=12
\end{array} \Rightarrow\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-12=0 \Rightarrow x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}, x_{1} x_{2}=\frac{4 m^{2}-12}{3+4 k^{2}}\right. \text {. }
$$

Thus, $\overrightarrow{A M} \cdot \overrightarrow{A N}=\left(x_{1}-2\right)\left(x_{2}-2\right)+\left(k x_{1}+m\right)\left(k x_{2}+m\right)$
$$
\begin{array}{l}
=\left(1+k^{2}\right) x_{1} x_{2}+(k m-2)\left(x_{1}+x_{2}\right)+m^{2}+4=0 \\
\Rightarrow\left(1+k^{2}\right)\left(4 m^{2}-12\right)-8 k m(k m-2)+\left(m^{2}+4\right)\left(3+4 k^{2}\right)=0 \\
\Rightarrow 7 m^{2}+16 k m+4 k^{2}=(7 m+2 k)(m+2 k)=0 .
\end{array}
$$

When $7 m+2 k=0$, $m=-\frac{2}{7} k \Rightarrow y=k x-\frac{2}{7} k=k\left(x-\frac{2}{7}\right)$,
at this time, the line $y=k x+m$ passes through the fixed point $\left(\frac{2}{7}, 0\right)$;
When $m+2 k=0$, $m=-2 k \Rightarrow y=k x-2 k=k(x-2)$,
at this time, the line $y=k x+m$ passes through the fixed point $A(2,0)$, which does not meet the requirement, so it is discarded.
In summary, the line $l$ passes through a fixed point, and the coordinates of the fixed point are $\left(\frac{2}{7}, 0\right)$.