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

Ellipse: $$x^2 + 9y^2 = 9$$, i.e., $$\frac{x^2}{9} + y^2 = 1$$.

Semi-major axis $$a = 3$$ (along x), semi-minor axis $$b = 1$$ (along y).

A = positive x-axis intersection = (3, 0). B = positive y-axis = (0, 1).

Circle C has major axis (length 6) as diameter: centre (0,0), radius 3. So $$C: x^2 + y^2 = 9$$.

Line through A(3,0) and B(0,1): $$\frac{x}{3} + y = 1$$ or $$x + 3y = 3$$.

Find intersection of $$x + 3y = 3$$ with $$x^2 + y^2 = 9$$:

$$x = 3 - 3y$$. Substituting: $$(3-3y)^2 + y^2 = 9$$

$$9 - 18y + 9y^2 + y^2 = 9$$

$$10y^2 - 18y = 0$$

$$y(10y - 18) = 0$$

$$y = 0$$ (point A) or $$y = 9/5$$

At $$y = 9/5$$: $$x = 3 - 27/5 = -12/5$$. So $$P = (-12/5, 9/5)$$.

Area of triangle OAP with O(0,0), A(3,0), P(-12/5, 9/5):

$$= \frac{1}{2}|x_A \cdot y_P - x_P \cdot y_A| = \frac{1}{2}|3 \cdot 9/5 - (-12/5) \cdot 0| = \frac{1}{2} \cdot \frac{27}{5} = \frac{27}{10}$$

$$m = 27, n = 10$$. Check coprime: gcd(27,10) = 1 ✓

$$m - n = 27 - 10 = 17$$

The correct answer is Option 3: 17.