First, trying to find the circle passing through $(0,1)$, $(0,-1)$ and $(a,0)$, where $-1\le a \le 1$ and $a\ne 0$. This is the circle which the meridian with longitude $90^\circ a$ would map to.
Let the circle have centre $(p,0)$ and radius $\sqrt{p^2 + 1}$:
$$\begin{align*}
(x-p)^2 + y^2 &= p^2 + 1\\
x^2 - 2px + p^2 + y^2 &= p^2 + 1\\
x^2 + y^2 - 2px - 1 &= 0
\end{align*}$$
Substitute $(x,y) = (a,0)$,
$$\begin{align*}
a^2 +0^2- 2pa -1 &= 0\\
-2ap &= 1-a^2\\
-2p &= \frac{1-a^2}{a}
\end{align*}$$
Then the equation of this circle, in terms of $a$, would be:
$$\begin{align*}
x^2 + y^2 + \frac{1-a^2}{a}x - 1 &= 0\\
ax^2 + ay^2 + (1-a^2)x - a &= 0 \tag 1\label 1
\end{align*}$$
If the circle of latitude $90^\circ b$ is mapped to a segment on $y=b$ (a degenerate circle), then to find the $x$-coordinate of the intersection of this line and the above circle:
$$\begin{align*}
&ax^2 + (1-a^2)x -a(1-b^2) = 0\\
x &= \frac{-(1-a^2) \pm \sqrt{(1-a^2)^2 - 4a[-a(1-b^2)]}}{2a}\\
&= \frac{-(1-a^2) \pm \sqrt{(1-a^2)^2 + 4a^2(1-b^2)}}{2a}\\
\end{align*}$$
Picking the $+$-root gives the $x$-coordinate that's within $[-1, 1]$ and has the same sign as $a$.
In terms of latitude $\phi$ and longitude $\lambda$ (relative to the centre meridian), where $\phi,\lambda\in [-\pi/2, \pi/2]$, the map from $(\phi, \lambda)$ to $(x,y)$ would be:
$$\begin{align*}
y &= \frac{\phi}{\pi/2}\\
x &= \begin{cases}
0, & \lambda = 0\\
\dfrac{-(1-(2\lambda/\pi)^2) + \sqrt{(1-(2\lambda/\pi)^2)^2 + 4(2\lambda/\pi)^2(1-(2\phi/\pi)^2)}}{2\cdot2\lambda/\pi},& \lambda \ne 0\\
\end{cases}\\
&= \begin{cases}
0, & \lambda = 0\\
\dfrac{-((\pi/2)^2-\lambda^2) + \sqrt{((\pi/2)^2-\lambda^2)^2 + 4\lambda^2((\pi/2)^2-\phi^2)}}{2\lambda\cdot \pi/2},& \lambda \ne 0\\
\end{cases}
\end{align*}$$
If the circle of latitude $90^\circ b$ is mapped to a circular arc that passes through both the $y$-axis and $x^2+y^2=1$ proportionally between $(0,1)$ and $(0,-1)$ (the two poles), then the arc would pass through
- $(0,b)$ on the $y$-axis;
- $\left(\cos \frac{b\pi}2, \sin \frac{b\pi}2\right)$ on the right semicircle; and
- $\left(-\cos \frac{b\pi}2, \sin \frac{b\pi}2\right)$ on the left semicircle.
According to WolframAlpha, the circle is centred at $\left(0, \frac{1-b^2}{2\sin \frac{b\pi}2 - 2b}\right)$, and the radius would be
$$\begin{align*}
\left|\frac{1-b^2}{2\sin \frac{b\pi}2 - 2b}-b\right|
&= \left|\frac{1 - 2b\sin \frac{b\pi}2 + b^2}{2\sin \frac{b\pi}2 - 2b}\right|\\
&= \left|\frac{\cos^2 \frac{b\pi}2 + \left(\sin \frac{b\pi}2 - b\right)^2}{2\left(\sin \frac{b\pi}2 - b\right)}\right|
\end{align*}$$
When $b\to\pm 1$, the $y$-coordinate of the centre tends to $\pm 1$, and the radius tends to $0$.
The equation of the circle, when $-1<b<1$ and $b\ne 0$, would be
$$x^2 + \left(y-\frac{1-b^2}{2\sin \frac{b\pi}2 - 2b}\right)^2
= \left(\frac{1 - 2b\sin \frac{b\pi}2 + b^2}{2\sin \frac{b\pi}2 - 2b}\right)^2 \tag2\label2$$
Then given latitude $90^\circ b$ and longitude $90^\circ a$ (relative to the centre meridian) (both non-$0^\circ$, and non-polar latitude), the mapped $(x,y)$ is the intersection of two circles $\eqref 1$ and $\eqref2$ on or inside the unit circle.
There are close forms of the intersection in terms of the two centres and radii.
