I'm working on a toy model for a supernova explosion.
The model is that there is a hot circular region at the north pole of an opaque sphere (with the circle centred on the pole).
The sphere has radius R and the circle radius r.
The sphere itself always fills $\pi R^2$
If you look directly down from N, the disc fills an area of $\pi r^2$
If you look from S, the sphere blocks the disc so it fills 0 area.
How do I work out how much projected area the disc fills for the angles in between?
On
You can consider the visible region of the sphere as a spherical cap (a portion of a sphere cut off by a circle drawn on the sphere). For simplicity, if you are viewing the sphere from very far away, we can assume the lines through all points of the surface in the direction of the viewer are parallel and that the visible region is a hemisphere.
The hot region also is a spherical cap. The visible part of the hot region is the intersection of two spherical caps, namely the hot region and the visible region.
Let $\theta$ be the angle between the north pole and the direction from which you are viewing the sphere. If the angle is not too large, you can see the entire hot region, as in the diagram below. In this diagram you are viewing the sphere from far away on the right side, and the thick red line represents the boundary of the hot region.
The projection of the hot region onto your "viewing plane" is the same as the projection of the disk of radius $r$ that is bounded by the red line. Of course you do not see points in the interior of the disk; you see points on the sphere's surface between you and the disk, but the points are on the same line in the direction from which you are viewing.
The disk projects to an ellipse. The long axis of the ellipse is $2r,$ the same as the diameter of the disk, but the short axis is foreshortened to $2r \cos\theta$ due to your viewing angle. So the area you see is
$$ r^2 \cos \theta. $$
This formula is good as long as the circle at the edge of the hot region does not intersect the great circle at the edge of the visible hemisphere. Let $a = \sqrt{R^2 - r^2},$ that is, $a$ is the distance from the center of the sphere to the disk of radius $r.$ Then as long as $\theta < \arctan \frac ar$ the circles do not intersect.
But if $\theta$ is slightly larger than $\arctan \frac ar$, however, the circles intersect, leading to the configuration shown in the diagram below.
The two circles intersect in a common chord that cuts a circular segment off the disk of radius $r$ under the hot region and also cuts a circular segment off the disk bounded by the great circle that borders the visible hemisphere. The visible part of the hot region consists of all the parts of the sphere between the viewer and the upper segment of the great circle, as well as all the parts of the sphere between the viewer and the lower part of the disk of radius $r$; that is, what you see is the projection of the two regions delimited by thick red lines in the figure above.
One of these regions, the upper segment of the great-circle disk, is illustrated on the left in the diagram below.
The area of this segment is $$ A_1 = R^2 \arcsin \frac bR - ab \csc \theta.$$
As we have already stated, $a = \sqrt{R^2 - r^2}$ in this formula. For $b$ we can use any of the following equations: \begin{align} b^2 &= R^2 - a^2 \csc^2 \theta \\ &= R^2 - (R^2 - r^2) (1 + \cot^2 \theta) \\ &= r^2 + r^2 \cot^2 \theta - R^2 \cot^2 \theta \\ &= r^2 - (R^2 - r^2) \cot^2 \theta. \end{align}
The area $A_1$ projects directly onto the viewing plane with no reduction in area.
Provided that $\arctan \frac ar < \theta < \frac\pi2,$ the area of the segment cut off from the disk of radius $r$ is $$ r^2 \arcsin \frac br - ab \cot \theta,$$
but we are interested in the lower part of the disk, which has area $$ A_2 = \pi r^2 - r^2 \arcsin \frac br + ab \cot \theta.$$
Moreover, since we are viewing it an an angle, the projected area is only $\cos\theta$ times the area of that part of the disk. The total projected area of the visible hot region therefore comes out to
$$ A_1 + A_2 \cos \theta.$$
For $\frac\pi2 < \theta < \pi - \arctan \frac ar$ the interesting part of the disk of radius $r$ is a segment cut off from the bottom, and the formula for $A_2$ becomes $$ A_2 = r^2 \arcsin \frac br + ab \cot \theta,$$ noting that in this case $\cot\theta < 0.$ The formula for $A_1$ is the same as before, and the projected area is still $$ A_1 + A_2 \cos \theta,$$ but now $\cos\theta < 0$ and the formula represents the fact that the projection of the area $A_2$ onto the visible plane actually represents a cold region of the sphere that is blocking our view of the part of the hot region represented by $A_1.$
For $\theta > \pi - \arctan \frac ar,$ we do not see the hot region at all.
I assume you are talking about the orthogonal projection. Let's denote the viewing angle from north pole as $\psi=0$ and from south pole as $\psi=\pi$.
If $\psi\le\pi/2$, sphere doesn't block the disk and we want to calculate only the projection area of a disk. In that case one axis of the ellipse will still be $r$ and another one will become $r\cos\psi$, so the total area is $A_e=\pi r^2\cos\psi$.
When$\psi>\pi/2$, sphere will block the disk, so we want to calculate the area of difference between the projection of a disk (ellipse $e$) and a projection of a sphere (circle $$c). This difference is $A_d=A_e-A_i$, where $A_i$ is the area of intersection of circle and disk. One can derive a formula for the later problem, which is complicated. And if you want one, please refer to this. The parameters (centre coordinates, radius/radii) of the circle are$(0,0), R$, the parameters of the ellipse are $(R\sin\psi,0), (r\cos\psi,r)$.
However there is a simpler solution if you are ok with a good estimate. You can generate slanted rays for many points of the disk and check, whether they are blocked by the sphere. For a point $\pmb p =(x_0,y_0,R)$ and the ray vector $\pmb n =(\sin\psi,0,\cos\psi)$, the distance to point $(0,0,0)$ is given by: $$ d^2 = |\pmb p - (\pmb p\cdot\pmb n)\pmb n|^2 = (x_0\cos\psi-R\sin\psi)^2+y_0^2. $$ You just need to calculate the fraction of rays for which $d^2>R^2$ and multiply it by the area of the ellipse $\pi r^2\cos\psi$. Just be sure to precompute the values for $\cos\psi$ and $\sin\psi$ before the loop, so the only calculations inside the loop are simple additions and multiplications.