1. Find the intercepts of the line

The line is $2x+3y-k=0, \, k>0.$

• On the $x$-axis, $y=0$: $2x-k=0 \Rightarrow x=\frac{k}{2}.$ So, $A\left(\frac{k}{2},0\right).$

• On the $y$-axis, $x=0$: $3y-k=0 \Rightarrow y=\frac{k}{3}.$ So, $B\left(0,\frac{k}{3}\right).$

2. Equation of the circle with diameter $AB$

For endpoints $(x_1,y_1)$ and $(x_2,y_2)$, the circle with diameter joining them is $\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0.$

Substitute $A\left(\frac{k}{2},0\right)$ and $B\left(0,\frac{k}{3}\right)$: $x\left(x-\frac{k}{2}\right)+y\left(y-\frac{k}{3}\right)=0.$

So, $x^2+y^2-\frac{k}{2}x-\frac{k}{3}y=0.$

Given circle is $x^2+y^2-3x-2y=0.$

Comparing coefficients: $\frac{k}{2}=3 \Rightarrow k=6,$ and also $\frac{k}{3}=2 \Rightarrow k=6,$ which is consistent.

Thus, $k=6.$

3. Write the ellipse

Given ellipse: $x^2+9y^2=k^2.$

Using $k=6$: $x^2+9y^2=36.$

Divide by $36$: $\frac{x^2}{36}+\frac{y^2}{4}=1.$

Hence, $a^2=36, \quad b^2=4,$ so $a=6,\quad b=2.$

4. Length of latus rectum of the ellipse

For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad (a>b),$ the length of the latus rectum is $\frac{2b^2}{a}.$

Therefore, $\text{latus rectum} = \frac{2\cdot 4}{6}=\frac{8}{6}=\frac{4}{3}.$

So, $\frac{m}{n}=\frac{4}{3},$ with coprime integers $m=4$, $n=3$.

5. Compute $2m+n$

$2m+n=2(4)+3=11.$

So the correct option is $\boxed{\text{C}}.$