I may be misunderstanding something, but it seems fairly obvious to me that the best circle is of diameter 3 cm. Here's my argument:
Start with a circle with diameter 0, and slowly increase it to a diameter of 1 cm. During this time, the circle only intersects with the inner ellipse. Thus, the "weighted area" (with the pluses and minuses as stated) goes from 0 cm2 to -π/4 cm2.
Next, increase the circle's diameter from 1 cm to 2 cm. During this time, the area of the outer elliptical annulus that is covered by the circle is increasing, as well as the area of the central ellipse that is covered. The behavior of the weighted area is harder to find exactly during this interval. However, for a diameter of 2 cm, the positive area covered is π/2 cm2, while the negative area covered is also π/2 cm2. Thus, the net weighted area is zero.
Finally, increase the circle's diameter from 2 cm to 3 cm. At this point, we are only covering "new" area within the elliptical annulus as we increase the diameter of the circle. This means that the weighted area is strictly increasing over this range of diameters. It will therefore necessarily be greatest at a diameter of 3 cm, and the weighted area will be the area of the annulus minus the area of the inner ellipse, or π/2 cm2.
EDIT: To show a little more rigorously that there is not a local maximum of the area in range of diameters between 1 cm and 2 cm, consider the new areas that are covered when the circle expands from radius $r$ to radius $r + dr$:
As the circle expands, it covers a thin new ring. Some of this area (colored blue in the diagram) will be "positive area" in the outer annulus, while some of this new area (in orange) will be "negative area". But in the limit of infinitesimally small $dr$, the area of each blue strip will be proportional to the arc length of the circle that already lies in the outer annulus, and the area of each orange strip will be proportional to the arc length of the circle that lies inside the inner ellipse. This then implies that the change in the "weighted area" will be proportional to the difference in the arc lengths (arc length in the outer annulus minus area in the inner ellipse):
$$
\frac{d A_\text{weighted}}{dr} \propto (\text{arc length in outer region}) - (\text{arc length in inner region}) = r \left[ (\text{angle subtended by arc in outer region}) - (\text{angle subtended by arc in inner region})) \right]
$$
If there is a local minimum or maximum in this interval, it will occur when these two arc lengths are equal. But then by definition we will have
$$
\frac{d^2 A_\text{weighted}}{dr} = \frac{d}{dr} \left[ r (\text{difference of subtended angles}) \right] = r \frac{d}{dr} \left[ (\text{difference of subtended angles}) \right],
$$
since the difference in the subtended angles is by definition zero at this point. Since the difference of the subtended angles increases monotonically as $r$ increases, we conclude that this extremum is a local minimum. Thus, $A_\text{weighted}(r)$ does not have any local maxima between $r = 1/2$ and $r = 1$. Since it is negative $r = 1/2$ and zero at $r = 1$, it is therefore negative on the entire interval.
As an aside, you can use this argument to show where the minimum weighted area occurs. At this point, the blue area and the orange area in the above diagram will be the same, which (given the symmetry of the situation) implies that each strip subtends 90°. Thus, it will occur when the circle, the inner ellipse, and the line $x = y$ all mutually intersect. A quick bit of algebra shows that the radius of the circle at this point is $r = \sqrt{2/5}.$
