Solvable equation for angle between points on a sphere?

Question

I'm looking for an equation that would give the coordinates (latitude $\delta$,longitude $\phi$) (on a sphere) for all the points on the surface of the sphere that have a certain angular distance $\Phi$ (angle centered on the center of the sphere) to a certain point of coordinates $\delta_0$,$\phi_0$.

Answer 1

Start with $\delta_0=90°$ (north pole): in that case the circle of angular distance $\Phi$ has parametric equations $$ \cases{ x=\cos t \sin \Phi, \\ y=\sin t \sin \Phi, \\ z=\cos \Phi. } $$ Perform then a rotation by $\pi/2-\delta_0$ about the $y$ axis, followed by a rotation by $\phi_0$ about the $z$ axis, to carry the center at latitude $\delta_0$ and longitude $\phi_0$. The parametric equations become: $$ \cases{ x=\cos\phi_0 \cos\delta_0\cos\Phi+\sin\delta_0\cos t\sin\Phi-\sin t\sin\Phi\sin\phi_0,\\ y=\sin\phi_0\cos\delta_0\cos\Phi+\sin\delta_0\cos t\sin\Phi+\sin t\sin\Phi\cos\phi_0,\\ z=\sin\delta_0\cos\Phi-\cos\delta_0\cos t\sin\Phi. } $$ Comparing that with the standard polar coordinates $$ \cases{ x=\cos\delta \cos\phi, \\ y=\cos\delta \sin \phi, \\ z=\sin\delta, } $$ one finally gets: $$ \cases{ \sin\delta=\sin\delta_0\cos\Phi-\cos\delta_0\cos t\sin\Phi\\ \\ \cos\delta=\sqrt{\left(\cos \delta _0\cos\Phi+\sin\delta_0\cos t\sin\Phi\right)^2 +\sin^2t \sin^2\Phi}\\ \\ \sin\phi=\frac{\displaystyle \sin\phi_0\cos\delta_0\cos\Phi+\sin\delta_0\cos t\sin\Phi+\sin t\sin\Phi\cos\phi_0} {\displaystyle\sqrt{\left(\cos \delta _0\cos\Phi+\sin\delta_0\cos t\sin\Phi\right)^2 +\sin^2t \sin^2\Phi}}\\ \\ \cos\phi=\frac{\displaystyle \cos\phi_0 \cos\delta_0\cos\Phi+\sin\delta_0\cos t\sin\Phi-\sin t\sin\Phi\sin\phi_0} {\displaystyle\sqrt{\left(\cos \delta _0\cos\Phi+\sin\delta_0\cos t\sin\Phi\right)^2 +\sin^2t \sin^2\Phi}}\\ } $$