13. (15 points) As shown in Figure 2, in the Cartesian coordinate system, the equation of circle $\odot M$ is
$$
x^{2}+y^{2}+D x+E y+F=0,
$$

and the quadrilateral $A B C D$ inscribed in $\odot M$ has diagonals $A C$ and $B D$ that are perpendicular to each other, with $A C$ and $B D$ lying on the $x$-axis and $y$-axis, respectively.
(1) Prove: $F<0$;
(2) If the area of quadrilateral $A B C D$ is 8, the length of diagonal $A C$ is 2, and $\overrightarrow{A B} \cdot \overrightarrow{A D}=0$, find the value of $D^{2}+E^{2}-4 F$.
13. (1) Let $A(a, 0), C(c, 0)$.

From the problem, points $A$ and $C$ are on the negative and positive halves of the $x$-axis, respectively. Thus, $a c<0$.
When $y=0$, the equation becomes
$$
x^{2}+D x+F=0 \text{, }
$$

where the two roots of the equation are the $x$-coordinates of points $A$ and $C$.
Therefore, $x_{1} x_{c}=a c=F<0$.
(2) From the problem, we know
$S_{\text {quadrilateral } A B C D}=\frac{|\overrightarrow{A C}||\overrightarrow{B D}|}{2}=8$.
Also, $|\overrightarrow{A C}|=2$, so $|\overrightarrow{B D}|=8$.
Since $\overrightarrow{A B} \cdot \overrightarrow{A D}=0$, we have $\angle A=90^{\circ}$.
Thus, $|\overrightarrow{B D}|=2 r=8 \Rightarrow r=4$.
From the circle represented by equation (1), we have
$$
\frac{D^{2}}{4}+\frac{E^{2}}{4}-F=r^{2} \text{. }
$$

Therefore, $D^{2}+E^{2}-4 F=4 r^{2}=64$.