Let the ellipse $E:x^2+9y^2=9$ intersect the positive $x$- and $y$-axes at the points $A$ and $B$ respectively. Let the major axis of $E$ be a diameter of the circle $C$. Let the line passing through $A$ and $B$ meet the circle $C$ at the point $P$. If the area of the triangle with vertices $A, P$ and the origin $O$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $m-n$ is equal to

Given,

Ellipse $E:x^2+9y^2=9 ........\left(\mathrm{i}\right)$

Now point $A\left(3,0\right) \& B\left(0,1\right)$ which is the intersection of given ellipse with positive axis,

So, equation of line passing through $A \& B$ is given by,

$L:\frac{x}{3}+\frac{y}{1}=1\Rightarrow x=3-3y .......\left(\mathrm{ii}\right)$

Now equation of Circle with diametric point $\left(-3,0\right) \& \left(3,0\right)$ is given by,

$C:x^2+y^2=9 ........\left(\mathrm{iii}\right)$

Let $Q$ be foot of perpendicular from $P$ upon major axis,

So, from $\left(\mathrm{ii}\right) \& \left(\mathrm{iii}\right)$ we get,

${\left(3-3y\right)}^2+y^2=9$

$\Rightarrow y=\frac{9}{5}, 0$

Hence, $PQ=\frac{9}{5}$

Now Area of triangle will be,

$=\frac{1}{2}\times OA\times PQ=\frac{1}{2}\times 3\times \frac{9}{5}=\frac{27}{10}$

Hence, $m-n=17$