Area of Triangle Formed with Ellipse Auxiliary Circle Intersection

To find the area of the triangle formed by $A$, $M$, and the origin $O$, we proceed as follows:

1. Equation of the Ellipse and Auxiliary Circle:
   The given equation of the ellipse is:
   $x^2 + 9y^2 = 9 \implies \frac{x^2}{3^2} + \frac{y^2}{1^2} = 1$
   Thus, the semi-major axis is $a = 3$ (along the $x$-axis) and the semi-minor axis is $b = 1$ (along the $y$-axis).

   The auxiliary circle of the ellipse is the circle described on the major axis as diameter, which has radius $a = 3$ and center at the origin $O(0, 0)$:
   $x^2 + y^2 = 3^2 = 9$

2. Extremities of the Axes:

   • Extremity of the major axis: $A(3, 0)$
   • Extremity of the minor axis: $B(0, 1)$

3. Equation of the Line $AB$:
   The equation of the straight line passing through $A(3, 0)$ and $B(0, 1)$ is:
   $\frac{x}{3} + \frac{y}{1} = 1 \implies x + 3y = 3 \implies x = 3(1 - y)$

4. Finding the Intersection Point $M$:
   Substitute $x = 3(1 - y)$ into the auxiliary circle equation $x^2 + y^2 = 9$:
   $[3(1 - y)]^2 + y^2 = 9$
   $9(1 - 2y + y^2) + y^2 = 9$
   $9 - 18y + 9y^2 + y^2 = 9$
   $10y^2 - 18y = 0 \implies 2y(5y - 9) = 0$

   This yields two solutions:

   • $y = 0 \implies x = 3$, which gives the point $A(3, 0)$.
   • $y = \frac{9}{5} \implies x = 3\left(1 - \frac{9}{5}\right) = -\frac{12}{5}$, which gives the point $M\left(-\frac{12}{5}, \frac{9}{5}\right)$.

5. Area of Triangle $OAM$:
   The vertices of the triangle are $O(0, 0)$, $A(3, 0)$, and $M\left(-\frac{12}{5}, \frac{9}{5}\right)$.

   Since the base $OA$ lies along the $x$-axis, its length is:
   $\text{Base } OA = 3$
   The height of the triangle corresponding to the base $OA$ is the absolute value of the $y$-coordinate of point $M$:
   $\text{Height } h = \left|\frac{9}{5}\right| = \frac{9}{5}$

   Therefore, the area of $\triangle OAM$ is:
   $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 3 \times \frac{9}{5} = \frac{27}{10}$

Hence, the correct option is D.