Here's a simple geometric derivation. You don't even need spherical trigonometry.
The following figure shows a cross-section of the sphere's orthographic projection.
the horizontal axis is the $y$ axis and the vertical axis is the $z$ axis.
In this figure, the point labeled $C$ is the center of the sphere.
The line $CN$ is the north-south axis of the sphere.
The point $P$ on the $z$ axis is at the latitude $\varphi_0$ identified with the center of the projection;
the orthographic projection you are trying to compute is a projection parallel to the $z$ axis.
The figure shown above is itself an orthographic projection, but it is projected along the $x$ axis rather than the $z$ axis so that all the lines of latitude are projected to straight line segments.
Suppose we take an arbitrary line of latitude projected onto the line segment $LL'$.
The latitude of this line of latitude is labeled $\varphi$.
The outer circle of the projection of the sphere in the $x,y$ plane is simply the intersection of the sphere with the $x,y$ plane, represented by the horizontal line in the figure.
The points (if any) where the line of latitude $\varphi$ touches this circle must lie in the $x,y$ plane and also in the plane of the line of latitude;
projected parallel the $x$ axis, these points (if they exist) are projected to the point $T$ where the line $LL'$ intersects the $y$ axis.
Let $M$ be the point where the line $LL'$ intersects the axis $CN$.
The triangle $\triangle CML$ is a right triangle with right angle at $M$ and hypotenuse $CL = r$.
Leg $CM$ is opposite the angle $\angle CLM = \varphi$, so by simple plane trigonometry, $CM = r\sin\varphi$.
The triangle $\triangle CMT$ is also a right triangle with right angle at $M$.
Since leg $CM = r\sin\phi$ is adjacent to angle $\angle MCT = \varphi_0$,
the hypotenuse is $CT = r\sin\varphi\sec\varphi_0$.
Therefore the points you are looking for have $y$-coordinate
$$ y = r\sin\varphi\sec\varphi_0 = \frac{r \sin\varphi}{\cos\varphi_0} $$
provided that $y \leq r$.
If $y > r$ the intersection point $T$ is outside the sphere, indicating that the segment $LL'$ is entirely on one side of the $x,y$ plane and the line of latitude $\varphi$ does not meet the boundary of the projection at all.
When $y \leq r$, since $x^2 + y^2 = r^2$ at the points of tangency of the ellipse and the circle, the $x$-coordinates of these points are
$$
x = \pm \sqrt{r^2 - y^2}
= \pm \sqrt{r^2 - \left(\frac{r \sin\varphi}{\cos\varphi_0}\right)^2}
= \pm r \sqrt{1 - \frac{\sin^2\varphi}{\cos^2\varphi_0}}.
$$
Note that when $\varphi < 0$, the value of $\sin\varphi$ is negative,
the point $M$ is below the $y$ axis, and the point $T$ is to the left of $C$,
so the $y$-coordinate of $T$ is still $r\sin\varphi\sec\varphi_0$
(a negative value, since it is left of the origin).
If $\varphi_0$ is negative, the point $N$ on the axis $CN$ is below the $y$ axis rather than above it, but $T$ is still at a positive $y$-coordinate if $\varphi > 0$
and negative if $\varphi < 0$.
So the formulas above are good for all values of $\varphi_0$ and $\varphi$, although the figure shows only the results for positive values of those angles.
If you want the explicit longitudes of the points at latitude $\varphi$
that project to the coordinates $x,y$ calculated above,
note that $ML = r\cos\varphi$ (the radius of the line of latitude)
and $MT = r\sin\varphi\tan\varphi_0$.
Assuming $P$ is at longitude $\lambda_0$,
the longitude $\lambda$ at a point of tangency will satisfy
$$
\cos(\lambda - \lambda_0)
= - \frac{r\sin\varphi\tan\varphi_0}{r\cos\varphi}
= - \tan\varphi\tan\varphi_0.
$$
Therefore
$$ \lambda = \lambda_0 \pm \arccos(- \tan\varphi\tan\varphi_0)
= \lambda_0 \pm (\pi - \arccos(\tan\varphi\tan\varphi_0)).
$$
