2 circles and one ellipse and minimum area problem.

Not sure if this helps any, but if you've solved for the $r_1=r_2$ case, I think you should be able to follow from this line of reasoning for the general case.

There exists a triangle with vertices at the centers of the two circles and a third vertex elsewhere. The side which joins the centers of the two circles is of length $r_1+r_2$ and the other two sides contain the radii of each circle pointed towards their respective intersections with the ellipse.

First, optimize the two base angles of this triangle (the angles at the center of each circle, let's call them $\theta_1$ and $\theta_2$) so that the tangents to the circles at each of the three intersections become altitudes of the triangle and so that their intersection is thus the orthocenter of the triangle. Obviously the ratio between $r_1$ and $r_2$ will limit what angles $\theta_1$ and $\theta_2$ can be allowed.

Next, tilt your ellipse so that its major axis is also pointed towards this orthocenter, namely so that its most convex point will overlap the triangle, thus reducing the greatest area possible.

The distances from each tangent point to the orthocenter are equal and each line from the orthocenter to the centers of the circles bisects their angles $\theta_1$ and $\theta_2$.

The segment that joins the two points of intersection of the ellipse is a chord perpendicular to and bisected by the major axis of the ellipse. The optimized solution will have a minimum length for this chord. The angle of elevation of this chord (and thus the minor axis of the ellipse itself) relative to the line connecting the two circles is given by:

$\alpha = \tan^{-1}(\frac{r_2\sin(\theta_2)-r_1\sin(\theta_1)}{r_2(1-\cos(\theta_2)) + r_1(1-\cos(\theta_1))})$

From there, knowing the tilt of the ellipse relative to the line connecting the centers of the two circles and knowing that the angle covering the ellipse's segment is bisected by the major axis, I think you should be able to proceed similarly to the case $r_1 = r_2$. Or at least that's my attempt at an answer before the bounty runs out. Best of luck.

EDIT:

Definitions:

$\theta_1$ is $m\angle CO_2B$

$\theta_2$ is $m\angle AO_1B$,

$\theta_3$ is $m\angle AOC$

(in general, subscript 3 will refer to the ellipse, even though it is not marked as such on the diagram)

Relations between angles:

$\theta_2 = 2\arctan(\frac{r_1}{r_2}\tan(\frac{\theta_1}{2}))$

$\theta_3 = 2\arcsin(\frac{r_1}{b}\tan(\frac{\theta_1}{2})\sin(\frac{\theta_1 + \theta_2}{2}))$

Areas of the three segments:

$A_{\bigcirc_1} = \frac{\theta_1}{2}r_1^2$

$A_{\bigcirc_2} = \frac{\theta_2}{2}r_2^2$

$A_{\bigcirc_3} = a\cdot b\cdot \arctan(\frac{b}{a}\tan(\frac{\theta_3}{2}))$

see this post on finding the area of an ellipse segment

Areas of the triangles that need to be subtracted from those segments:

$A_{\triangle_1} = \frac{1}{2}r_1^2\sin(\theta_1)$

$A_{\triangle_2} = \frac{1}{2}r_2^2\sin(\theta_2)$

$A_{\triangle_3} = \frac{1}{2}(b^2\sin^2(\frac{\theta_3}{2}) + a^2\cos^2(\frac{\theta_3}{2}))\cdot\sin(\theta_3)$

You can use Heron's formula to calculate the area of $\triangle ABC$ using side-lengths defined by

$\overline{AB} = 2r_1\sin(\frac{\theta_1}{1})$

$\overline{BC} = 2r_2\sin(\frac{\theta_2}{2})$

$\overline{AC} = 2b\cdot\sin(\frac{\theta_3}{2})$

The constraint on tangents:

So everything above gives you results only in terms of your constants and the angles, which can all be reduced to being relations of a single angle of your choice. What you need to do to get a solution, is to remember the constraint that the tangent lines of your ellipse need to be the same as your tangent lines to your circle so that the ellipse never overlaps them.

The slope of the tangent lines to your ellipse is given by:

$s_{ellipse} = \pm\frac{b}{a}\cot(\frac{a}{b}\arctan(\frac{\theta_3}{2}))$

As for the circles, you can think of $\alpha$ as defined in my original answer as the circles' rotation away from the $y$-axis, which means that for circle 1 the measure of the angle away from the positive $x$ direction is:

$\theta_1^\prime = -(\theta_1 + \frac{\pi}{2} - \alpha)$

And for circle 2 we get:

$\theta_2^\prime = \theta_2 + \frac{\pi}{2} + \alpha$

Your slopes for these tangents are then given by:

$s_1 = -\cot(\theta_1^\prime)$

$s_2 = -\cot(\theta_2^\prime)$

Setting those slopes equal to the relevant (positive or negative) version of the ellipse's slope and rewriting your angles in terms of a single angle using the relationships given right at the beginning of these edits, should give you a solution for your angle, which then becomes a solution for all three angles, when then should give you a solution for your area... hopefully I haven't made any typos.