13. (20 points) Two circles on the same side of the $x$-axis: a moving circle $C_{1}$ and the circle $4 a^{2} x^{2}+4 a^{2} y^{2}-4 a b x-2 a y+b^{2}=0$ are externally tangent $(a, b \in \mathbf{N}, a \neq 0)$, and the moving circle $C_{1}$ is tangent to the $x$-axis. Find
(1) the equation of the locus $\Gamma$ of the center of the moving circle $C_{1}$;
(2) if the line $4(\sqrt{7}-1) a b x-4 a y+b^{2}+a^{2}-6958 a=0$ intersects the curve $\Gamma$ at exactly one point, find the values of $a$ and $b$.
3,13. (1) From $4 a^{2} x^{2}+4 a^{2} y^{2}-4 a b x-2 a y+b^{2}=0$, we can get
$$
\left(x-\frac{b}{2 a}\right)^{2}+\left(y-\frac{1}{4 a}\right)^{2}=\left(\frac{1}{4 a}\right)^{2} .
$$

Given $a, b \in \mathbf{N}$ and the two circles are on the same side of the $x$-axis, it is known that the center of the moving circle is above the $x$-axis. Let the coordinates of the center of the moving circle be $(x, y)$, then we have
$$
\sqrt{\left(x-\frac{b}{2 a}\right)^{2}+\left(y-\frac{1}{4 a}\right)^{2}}=y+\frac{1}{4 a} \text {. }
$$

Rearranging, we get the equation of the trajectory of the center of the moving circle as
$$
y=a x^{2}-b x+\frac{b^{2}}{4 a}\left(x \neq \frac{b}{2 a}\right) .
$$

Another solution: From the given, the trajectory of the center of the moving circle is a parabola with focus at $\left(\frac{b}{2 a}, \frac{1}{4 a}\right)$, directrix at $y=-\frac{1}{4 a}$, and vertex at $\left(\frac{b}{2 a}, 0\right)$ (excluding this point). Thus, the trajectory equation is
$$
\left(x-\frac{b}{2 a}\right)^{2}=\frac{1}{a} y,
$$

which is $y=a x^{2}-b x+\frac{b^{2}}{4 a}\left(x \neq \frac{b}{2 a}\right)$.
(2) Solving the system of equations
$$
\left\{\begin{array}{l}
y=a x^{2}-b x+\frac{b^{2}}{4 a}\left(x \neq \frac{b}{2 a}\right), \\
4(\sqrt{7}-1) a b x-4 a y+b^{2}+a^{2}-6958 a=0 .
\end{array}\right.
$$

Eliminating $y$ gives $4 a^{2} x^{2}-4 \sqrt{7} a b x-\left(a^{2}-6958 a\right)=0$.
From $\Delta=16 \times 7 a^{2} b^{2}+16 a^{2}\left(a^{2}-6958 a\right)=0$, we get $7 b^{2}+a^{2}=6958 a$.
From equation (3), we know $71 a^{2}$, thus, $71 a$.
Let $a=7 a_{1}$, substituting into equation (3) gives $b^{2}+7 a_{1}^{2}=6958 a_{1}$.
Then $7 \mid b^{2}$, thus, $7 \mid b$.
Let $b=7 b_{1}$, substituting into the above equation gives $7 b_{1}^{2}+a_{1}^{2}=994 a_{1}$.
Similarly, $7\left|a_{1}, 7\right| b_{1}$.
Let $a=49 n, b=49 m$, substituting into equation (3) gives
$7 m^{2}+n^{2}=142 n$.
Completing the square for equation (1), we get $(n-71)^{2}+7 m^{2}=71^{2}$.
Analyzing the parity of the above equation, we know that $m, n$ are both even, so $7 m^{2}=71^{2}-(n-71)^{2}$ is a multiple of 8. Hence $41 m$.
Let $m=4 r$, then $112 r^{2} \leqslant 71^{2}$. Thus, $r^{2} \leqslant 45$.
Therefore, $|r|=0,1,2,3,4,5,6$.
Only when $|r|=0,4$, $71^{2}-112 r^{2}$ is a perfect square. Thus, the solutions are
$$
\left\{\begin{array} { l } 
{ a = 6 9 5 8 , } \\
{ b = 0 \text { (discard) } ; }
\end{array} \left\{\begin{array} { l } 
{ a = 6 2 7 2 } \\
{ b = 7 8 4 ; }
\end{array} \left\{\begin{array}{l}
a=686, \\
b=784 .
\end{array}\right.\right.\right.
$$