1. Write the ellipse in standard form

Given $x^2+9y^2=9$ Divide by $9$: $\frac{x^2}{9}+\frac{y^2}{1}=1$ So the ellipse has semi-major axis $a=3$ along the $x$-axis and semi-minor axis $b=1$ along the $y$-axis.

Hence:

• point on positive $x$-axis: $A=(3,0)$
• point on positive $y$-axis: $B=(0,1)$

2. Circle having major axis of ellipse as diameter

The major axis of the ellipse is the segment from $(-3,0)$ to $(3,0)$, whose length is $6$. So the circle $C$ has diameter $6$, center at origin, and radius $3$.

Therefore its equation is $x^2+y^2=9$

3. Equation of line through $A$ and $B$

Through $A=(3,0)$ and $B=(0,1)$, slope is $m=\frac{1-0}{0-3}=-\frac13$ So the line is $y=-\frac13(x-3)= -\frac{x}{3}+1$ That is, $x+3y=3$

4. Find the second intersection point $P$ of the line with the circle

Since $A$ already lies on the circle, the other intersection is $P$. Substitute $y=1-\frac{x}{3}$ into $x^2+y^2=9$

Then $x^2+\left(1-\frac{x}{3}\right)^2=9$ $x^2+1-\frac{2x}{3}+\frac{x^2}{9}=9$ Multiply by $9$: $9x^2+9-6x+x^2=81$ $10x^2-6x-72=0$ $5x^2-3x-36=0$

Solve: $5x^2-3x-36=(x-3)(5x+12)=0$ So $x=3 \quad \text{or} \quad x=-\frac{12}{5}$ The value $x=3$ gives point $A$, hence $x_P=-\frac{12}{5}$ Now $y_P=1-\frac{1}{3}\left(-\frac{12}{5}\right)=1+\frac{4}{5}=\frac{9}{5}$ Thus $P=\left(-\frac{12}{5},\frac{9}{5}\right)$

5. Area of triangle $AOP$

Vertices are $O=(0,0),\quad A=(3,0),\quad P=\left(-\frac{12}{5},\frac{9}{5}\right)$

Using determinant formula: $\text{Area}=\frac12\left|x_Ay_P-y_Ax_P\right|$ $=\frac12\left|3\cdot \frac95-0\cdot\left(-\frac{12}{5}\right)\right|$ $=\frac12\cdot \frac{27}{5}=\frac{27}{10}$

So $\frac{m}{n}=\frac{27}{10}$ with coprime integers $m=27$, $n=10$. Hence $m-n=27-10=17$

6. Compare with options

The correct option is: $\boxed{17}$ which is Option C.