17. (18 points) Given the ellipse
$$
C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

has an eccentricity of $\frac{\sqrt{3}}{2}$, and the right focus is the center of the circle
$$
C_{2}:(x-\sqrt{3})^{2}+y^{2}=7
$$

(1) Find the equation of the ellipse $C_{1}$;
(2) If the line $l$ intersects the curves $C_{1}$ and $C_{2}$ at only one point each, and the intersection point of line $l$ and circle $C_{2}$ is $A$, find the coordinates of point $A$.
17. (1) Let the semi-focal length of the ellipse $C_{1}$ be $c$.
Then $\left\{\begin{array}{l}c=\sqrt{3}, \\ \frac{c}{a}=\frac{\sqrt{3}}{2}\end{array} \Rightarrow\left\{\begin{array}{l}a=2, \\ b=1 .\end{array}\right.\right.$
Thus, the equation of the ellipse is $\frac{x^{2}}{4}+y^{2}=1$.
(2) When the slope of the line $l$ does not exist, it clearly does not satisfy the problem.

When the slope of the line $l$ exists, let the equation of the line $l$ be $y=k x+m(k, m \in \mathbf{R})$, and the point $A\left(x_{A}, y_{A}\right)$.
Solving the system of equations $\left\{\begin{array}{l}\frac{x^{2}}{4}+y^{2}=1, \\ y=k x+m,\end{array}\right.$ and eliminating $y$ yields
$$
\left(1+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-4=0 \text {. }
$$

Thus, $\Delta_{1}=16\left(4 k^{2}-m^{2}+1\right)=0$, which means
$4 k^{2}-m^{2}+1=0$.
Solving the system of equations $\left\{\begin{array}{l}(x-\sqrt{3})^{2}+y^{2}=7, \\ y=k x+m,\end{array}\right.$ and eliminating $y$ yields
$$
\left(1+k^{2}\right) x^{2}+2(k m-\sqrt{3}) x+m^{2}-4=0 \text {. }
$$

Thus, $\Delta_{2}=16\left(4 k^{2}-m^{2}-2 \sqrt{3} m k+7\right)=0$, which means
$4 k^{2}-m^{2}-2 \sqrt{3} m k+7=0$.
Subtracting (2) from (4) gives $k m=\sqrt{3}$.

Substituting the above equation into equation (3) yields
$$
x_{A}=-\frac{k m-\sqrt{3}}{1+k^{2}}=0 \text {. }
$$

Substituting the above equation into the circle $C_{2}$ gives $y_{A}= \pm 2$.
Upon verification, $A(0,2)$ and $A(0,-2)$ satisfy the problem.
Thus, the points $A(0,2)$ and $A(0,-2)$.