Area of a circle as it changes to an ellipse when viewing at different angles
As you’ve observed, the projected line segment is not symmetric about the view line to the circle’s center—the center shifts. Another reason that you might be having some trouble working this out is that the apparent width of the ellipse also depends on the visual angle subtended by the circle, which varies with both the distance of the viewpoint from the center of the circle and the circle’s radius.
(Diagram provided by the OP.)
This problem can be reduced to finding the intersection of pairs of lines, which is easily done using homogeneous coordinates. We place the circle on the $x$-$y$ plane, centered at the origin, and Let the viewpoint be $V=(0,d\cos\theta,d\sin\theta)$, ($d>0$). We’ll first look at what’s going on in the $y$-$z$ plane. In the above diagram, the horizontal direction represents $z$ and the vertical $y$. To reduce clutter, the $x$-coordinate will be suppressed and, at the risk of introducing some confusion, the others will be designated $x'$ and $y'$, respectively.
The line perpendicular to $\overline{OV}$ that passes through the upper point $A(0,r)$ is $y'=r-x'\cot\theta$, which we can rewrite in normal form as $x'\cos\theta+y'\sin\theta-r\sin\theta=0$. The homogeneous vector that represents this line consists of the coefficients of the latter equation: $$\mathbf l=(\cos\theta,\sin\theta,-r\sin\theta).$$ The line through $V$ and the lower point $B(0,-r)$ is represented by the cross product of the homogeneous coordinates of these points: $$\mathbf m=(d\cos\theta,d\sin\theta,1)\times(0,-r,1)=(r+d\sin\theta,-d\cos\theta,-dr\cos\theta).$$ Their intersection is $$\mathbf l\times\mathbf m=(-dr\sin2\theta,dr\cos2\theta-r^2\sin\theta,-(d+r\sin\theta))$$ which in Cartesian coordinates is $$C=\left({dr\sin2\theta\over d+r\sin\theta},{r^2\sin\theta-dr\cos2\theta\over d+r\sin\theta}\right).$$ The minor axis length of the circle’s image is $AC$, which you can compute using the standard formula for the distance between two points as $${2dr\cos\theta\over d+r\sin\theta}.$$ Compared to your initial guess, there’s an extra factor of ${d\over d+r\sin\theta}$ that accounts for the asymmetry of the view and the “angular size correction” factor. As $d\to\infty$, so that the projection becomes closer and closer to parallel, this correction factor approaches unity.
This isn’t the whole story, though. Rays from the viewpoint through points on the circle converge as they get closer to the viewpoint, so the other semi-axis of the circle’s projection isn’t going to be equal to $r$, either. To work out what this is, first project the center of the ellipse back to the $x$-$y$ plane. This center is the midpoint of $A$ and $C$ and its pre-image can be found by intersecting lines again: $$\begin{align}D &= \frac12\left((0,r)+\left({dr\sin2\theta\over d+r\sin\theta},{r^2\sin\theta-dr\cos2\theta\over d+r\sin\theta}\right)\right) \\ &=\left(\frac{d r \sin (\theta ) \cos (\theta )}{d+r \sin (\theta )},\frac{r \sin (\theta ) (d \sin (\theta )+r)}{d+r \sin (\theta )}\right)\end{align}$$ and its back-projection is $$\overline{VD}\times\overline{OA}=(V\times D)\times(-1,0,0)=\left(0,\frac{d r^2 \sin (\theta ) \cos (\theta )}{d+r \sin (\theta )},\frac{d^2 \cos (\theta )}{d+r \sin (\theta )}\right)$$ which becomes $\left(0,\frac{r^2}d\sin\theta\right)$ in Cartesian coordinates.
Going back to 3-D coordinates, this is the point $\left(0,\frac{r^2}d\sin\theta,0\right)$. The $x$-coordinates of the points on the circle with this $y$-coordinate are $\pm\sqrt{r^2-y^2}=\pm r\sqrt{1-\left(\frac rd\sin\theta\right)^2}$. I’ll leave finding the projections of these points and the resulting semi-axis length to you (hint: you can use similar triangles).