14. Given the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{m}=1(m>0)$ passes through the point $A\left(1, \frac{3}{2}\right)$. Can we find two points $E, F$ on the ellipse $C$ such that: the slopes of the three sides of $\triangle A E F$ form an arithmetic sequence, and the lines containing sides $A E$ and $A F$ are both tangent to the circle $(x-2)^{2}+\left(y-\frac{3}{2}\right)^{2}=r^{2}(r>0)$? If they exist, find the corresponding value of $r$; if not, explain the reason.
Substitute point $A$ to get $\frac{1}{4}+\frac{9}{4 m}=1 \Rightarrow m=3$, then $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
As shown in the figure, let the equation of the tangent line be $y=k(x-1)+\frac{3}{2}$, then $\frac{|k|}{\sqrt{k^{2}+1}}=r \Rightarrow k^{2}=\frac{r^{2}}{1-r^{2}}$, thus $k_{A E}+k_{A F}=0$. From the graph, we know that the slopes of the three sides of $\triangle A E F$ form
an arithmetic sequence $\Rightarrow k_{E F}=3 k_{A E}$.
Solving the system $\left\{\begin{array}{l}y=k(x-1)+\frac{3}{2}, \\ 3 x^{2}+4 y^{2}=12\end{array} \Rightarrow\right.$ $\left(4 k^{2}+3\right) x^{2}-4 k(2 k-3) x+4 k^{2}-12 k-3=0$.
Thus, $x_{E}=\frac{4 k^{2}-12 k-3}{4 k^{2}+3}$, substituting $-k$ for $k$, we get $x_{F}=\frac{4 k^{2}+12 k-3}{4 k^{2}+3}$, hence $k_{E F}=\frac{k\left(x_{E}-1\right)+k\left(x_{F}-1\right)}{x_{E}-x_{F}}=\frac{k\left(8 k^{2}-6\right)-2 k\left(4 k^{2}+3\right)}{-24 k}=\frac{1}{2}$. Therefore, $k_{A E}=\frac{1}{6} \Rightarrow \frac{r^{2}}{1-r^{2}}=\frac{1}{36} \Rightarrow r=\frac{\sqrt{37}}{37}$.