<s>[INST] Distance between two circles in a sphere I have a sphere with radius $R$ and $O$ is the origin.
Inside sphere there are 3 circles. The small circle in black colour is fixed with it's position defined by and $\alpha$ angle and among other two circles one is great circle and another small circle can rotate by maintaining same distance and they are always parallel to each other. Both blue circles intersect the black great circle at point A and B. The rectilinear distance of A to B can be calculated using this relation:
$AB = 2RSin {\theta\over2}$ 
Orientation of upper blue great circle is defined as $\beta$ which is the angle starting from $z$ axis.The range of $\beta $ can vary from $0^0 -360^0$ and it is a known value. Though my both blue coloured circle is parallel to each other, is it possible to get the lower circle's orientation angle  from the centre of sphere that respect $\theta$ as it is the rectilinear distance? We always assume that both blue circle intersect the black circle.
 [/INST] The black circle is the intersection of a plane $P\cdot (0,0,1)=a$ (I don't think you've specified $a$) and the sphere $P\cdot P =1$.
The blue great circle is the intersection of $P\cdot (\cos \beta, 0, \sin \beta) = 0$ with the sphere, and the blue small circle is the intersection of $P\cdot (\cos \beta, 0, \sin \beta) = b$ with the sphere.
The line of intersection between $P\cdot (0,0,1)=a$ and $P\cdot (\cos \beta, 0, \sin \beta) = b$ is fairly obviously $$P(t) = (\frac{b - a\sin\beta}{\cos\beta}, t, a)$$. Intersection with the sphere gives $$t^2 = 1 - a^2 - \left(  \frac{b - a\sin\beta}{\cos\beta} \right)^2$$
The angle between two points on the unit sphere is the arccos of their dot product, so the equation to solve is $$\frac{0 - a\sin\beta}{\cos\beta} \frac{b - a\sin\beta}{\cos\beta} + \sqrt{ \left( 1 - a^2 - \left(  \frac{0 - a\sin\beta}{\cos\beta} \right)^2 \right) \left( 1 - a^2 - \left(  \frac{b - a\sin\beta}{\cos\beta} \right)^2 \right) } + a^2 = \cos \theta$$ which can be simplified to
$$\frac{- a\sin\beta(b - a\sin\beta) + \sqrt{(\cos^2\beta - a^2)(\cos^2\beta + 2ab\sin\beta - a^2 -b^2)}}{\cos^2 \beta} + a^2 = \cos \theta$$
which can be further rearranged into a quartic in $\sin\beta$ (WARNING: I may have made some errors while rearranging):
$$(1 - \cos^2\theta)\sin^4\beta - 2ab(1 - \cos \theta)(\sin^3\beta - \sin\beta) + (\cos^2\theta -k - 2) \sin^2\beta + ( k + 1) = 0$$ where $k = a^2 b^2 - 2a^2 - b^2 + 2a^2 \cos \theta - \cos^2\theta$
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