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I've found a formula for defining a great circle (since it's the set of points $(\theta, \varphi)$ such that their distance is $\pi/2$ from a given point $(\theta_0, \varphi_0)$):

$-tan(\varphi)tan(\varphi_0) = cos(\theta_0 - \theta)$

Now, I have two points on the sphere $(\theta_1,\varphi_1),(\theta_2,\varphi_2)$. I want to use them to define a great circle, but to do that, I need to use those two points in some way to derive $(\theta_0, \varphi_0)$ (either one of the two possibilities assuming the two points aren't antipodal or equal). How can I do so (whether using the above formula or another part of spherical geometry)?

Note: I want to do this entirely with spherical geometry and spherical coordinates if possible. I know that there's a very easy way to do it: simply take the cross-product of the 3D vector representation of the two points and normalize it. But I'm also doing a lot of other things in spherical coordinates and so hopefully a spherical geometry solution will help me with other similar problems.

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If $\varphi_1=\pm\frac\pi2,$ if $\varphi_1=\pm\frac\pi2,$ or if $\theta_1 - \theta_2$ is an integer multiple of $\pi$ then you know $\varphi_0=0$ and you can easily find $\theta_0$ if it is determined. If $\varphi_1=0$ (or if $\varphi_2=0$) then you know that $\theta_0 = \theta_1 \pm \frac\pi2$ (or $\theta_0 = \theta_2 \pm \frac\pi2,$ respectively) and can then easily find $\varphi_0$ if it is determined.

Now let's assume a case in which none of the conditions above is true. This implies that $\tan\varphi_0\neq 0.$

You know that if your great circle's pole is at $(\theta_0,\varphi_0)$ then your two points must satisfy the equations

\begin{align} -\tan\varphi_1\tan\varphi_0 &= \cos(\theta_0 - \theta_1) = \cos\theta_0\cos\theta_1 + \sin\theta_0\sin\theta_1, \tag1\\ -\tan\varphi_2\tan\varphi_0 &= \cos(\theta_0 - \theta_2) = \cos\theta_0\cos\theta_2 + \sin\theta_0\sin\theta_2. \tag2 \end{align}

Cross-multiply Equation $(1)$ and Equation $(2)$ to get $$ -\tan\varphi_1\tan\varphi_0 (\cos\theta_0\cos\theta_2 + \sin\theta_0\sin\theta_2) = -\tan\varphi_2\tan\varphi_0 (\cos\theta_0\cos\theta_1 + \sin\theta_0\sin\theta_1). $$

Since $\tan\varphi_0\neq 0,$ divide by $-\tan\varphi_0$ on both sides and distribute the multiplication to get $$ \tan\varphi_1 \cos\theta_0\cos\theta_2 + \tan\varphi_1 \sin\theta_0\sin\theta_2 = \tan\varphi_2 \cos\theta_0\cos\theta_1 + \tan\varphi_2 \sin\theta_0\sin\theta_1. $$

Collect terms in $\cos_0$ and $\sin_0$: $$ (\tan\varphi_1 \cos\theta_2 - \tan\varphi_2 \cos\theta_1) \cos\theta_0 = (\tan\varphi_2 \sin\theta_1 - \tan\varphi_1 \sin\theta_2) \sin\theta_0. $$

Since $\varphi_1,$ $\varphi_2,$ $\theta_1,$ and $\theta_2$ are all known, you now have an equation of the form $k_1 \cos\theta_0 = k_2 \sin\theta_0$ for known $k_1$ and $k_2$ and you can solve for $\theta_0.$ For example, if $\tan\varphi_2 \sin\theta_1 - \tan\varphi_1 \sin\theta_2 \neq 0$ then you can set $$ \theta_0 = \arctan\left(\frac{\tan\varphi_1 \cos\theta_2 - \tan\varphi_2 \cos\theta_1} {\tan\varphi_2 \sin\theta_1 - \tan\varphi_1 \sin\theta_2}\right). $$

Once you know $\theta_0$ you can use it to solve for $\varphi_0$ in one of equations $(1)$ or $(2).$

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