"Inclination" here seems to be connected with the angle $i$ at which a cone is intercepted by a plane to produce the ellipse, but I'm not familiar with the term. At any rate the OP says (see Comment above) that the semi-minor axis can be expressed as $b = a \cos i$. In what follows, we assume this has been done.
Confusion arises from the subsequent wording that "the ellipse semi-major and semi-minor axes are not collinear with the X- and Y- axes but are inclined by the position angle." Notation for this position angle is not provided in the Question, so let's supply it here, and in doing so let's refer instead to it as the angle of rotation $\alpha$. Specifically, since the ellipse is "centered on the origin", we define $\alpha \in [-\pi/2,\pi/2]$ to be the angle between the semi-major axis and the positive X-axis.
More concretely: the endpoints of the major axis, centered as they are on the origin, are expressed by:
$$ \pm a (\cos \alpha, \sin \alpha) $$
The first step is then "unrotation" by this angle of rotation, a linear transformation on Cartesian coordinates expressed by the following matrix multiplication:
$$ \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}
\begin{bmatrix} X \\ Y \end{bmatrix} $$
Clearly applying this transformation to the endpoints of the major axis (see above) will map them to $( \pm a, 0)$, thereby putting both major and minor axes in line with the $X$- and $Y$-coordinate axes, respectively.
In this "standard position" the ellipse can easily be transformed into a circle by scaling the X- or Y-coordinate or both. E.g. the mapping $(X,Y) \rightarrow (bX,aY)$ will produce a circle of radius $ab$ by scaling both coordinates. Combining these operations gives a matrix multiplication:
$$ \begin{bmatrix} X' \\ Y' \end{bmatrix} =
\begin{bmatrix} b\cos \alpha & b\sin \alpha \\ -a\sin \alpha & a\cos \alpha
\end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix} $$
Optional: Further equal scaling of both coordinates by a nonzero factor will change the radius accordingly (but preserve circularity) and might be desirable if a particular radius is desired.
Finally this circle may itself be rotated around the origin through any angle, using a linear transformation similar to the "unrotation" described above. However this step is unnecessary if the goal was merely to map the original ellipse to a circle.
Observe that an ellipse, with minor axis $a$ and major axis $b$, can be view as the part of the plane $z=y\tan\beta $ inside the cylinder $x^2+y^2=a^2$, where $\beta$ is the angle it forms with the $xy$-plane, satisfying $\cos\beta = a/b$.
Now, we could consider that the ellipse is made up from lots of ellipse rings that, when projected onto the $xy$-plane, become corresponding circles. Those circles can then be integrated easily.
The surface integral for the ellipse is then given by
$$ S=\int_0^a \int_0^{2\pi} f(r,\theta) rdr d\theta$$
where the projection, or the scaling, factor is actually fairly simple,
$$ f(r,\theta)=\sqrt{1+\left(z_y^{’}\right)^2}=\sec\beta= \frac{b}{a}$$
The area of the ellipse becomes
$$ S=\int_0^a \frac{2\pi b}{a}rdr \tag{1}$$
As seen, the area of the ellipse rings that fill up the whole ellipse is scaled as $2\pi r(b/a)$. This may also be viewed equivalently as the ‘circumference’ of each ellipse ring.
As expected, the surface integral (1) yields
$$S=\pi ab$$
Best Answer
Well, we know that the equation of a circle is given by:
$$\left(x-\text{a}\right)^2+\left(\text{y}-\text{b}\right)^2=\text{r}^2\tag1$$
Where $\left(\text{a},\text{b}\right)$ are the center coordinates of the circle and $\text{r}$ is the radius of the circle.
In your case, we have $\text{a}=\text{b}=5000$ and $\text{r}=2000$. So:
$$\left(x-5000\right)^2+\left(\text{y}-5000\right)^2=2000^2\tag2$$
We know that the equation of an ellipse is given by:
$$\left(\frac{x-x_0}{\alpha}\right)^2+\left(\frac{\text{y}-\text{y}_0}{\beta}\right)^2=1\tag3$$
Where $\left(x_0,\text{y}_0\right)$ are the center coordinates of the ellipse and $\alpha$ is the semi-major axis and $\beta$ is the semi-minor axis.
In your case, we have $x_0=2500$, $\text{y}_0=5000$, $\alpha=2000$, and $\beta=1000$. So:
$$\left(\frac{x-2500}{2000}\right)^2+\left(\frac{\text{y}-5000}{1000}\right)^2=1\tag4$$
Now, I used Mathematica to plot this with the following code:
And got the following output:
We can solve for the intersection points, using:
Using gridlines we can use the following code:
To see:
Now, it is not hard to show that the desired area is given by:
$$\mathcal{A}:=\text{I}_1+\text{I}_2\tag5$$
Where:
Where $\tau=\frac{500}{3}\left(35-2\sqrt{61}\right)$.
So, we get:
$$\mathcal{A}\approx2.00371\cdot10^6\tag8$$
And the exact value is: