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I have a sphere with radius $R$ and $O$ is the origin.enter image description here

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.

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    $\begingroup$ These are not all great circles. A great circle is the intersection of the sphere with a plane through the sphere's center. Consequently, all great circles have the same diameter -- the diameter of the sphere. Two distinct great circles necessarily intersect in two points. $\endgroup$ Jul 24, 2019 at 22:36
  • $\begingroup$ Is there any other name for those? I can edit. $\endgroup$
    – T. an
    Jul 24, 2019 at 22:40
  • $\begingroup$ A circle of a sphere is a circle lying on a sphere. The circles of a sphere come in two types: great circles, described previously, and small circles, which lie on any other plane and therefore have smaller radii than that of the sphere. $\endgroup$ Jul 24, 2019 at 23:18
  • $\begingroup$ @EricTowers Thanks you sir. I am editing according to that. $\endgroup$
    – T. an
    Jul 24, 2019 at 23:22
  • $\begingroup$ Shouldn't be $AB=2R \sin \theta/2$ ? $\endgroup$
    – Jean Marie
    Jul 25, 2019 at 3:05

2 Answers 2

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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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According to the comments on the question, I will start by assuming that $\alpha$ and $\beta$ are given.

Let's name some additional points on the sphere. Let $C$ be the point $(0,0,R)$ where the positive $z$ axis intersects the sphere. The blue great circle intersects the $y,z$ plane in two points; from those two points choose the one closer to $A$ and call it $D.$

The arrangement of points and arcs is shown in a schematic arrangement below. (This is not exactly to scale, because it is drawn as a plane figure while the actual figure is on the surface of a sphere.)

enter image description here

The figure below shows what it looks like on the sphere. Note that all arcs that are labeled with symbols ($\alpha,$ $\beta,$ $\gamma,$ $\delta,$ and $\theta$) are great-circle arcs, and the symbol is the name of the angle subtended by that arc at the center of the sphere.

enter image description here

Now we have a spherical triangle $\triangle ADC.$ Following the usual convention where the "sides" of the spherical triangle are measured by the angle (in radians) that they subtend at the center of the sphere, side $AC$ has measure $\alpha$ and side $CD$ has measure $\beta.$ The spherical angle $\angle ADC$ is $\frac\pi2.$ Let $\eta = \angle CAD.$ According to the spherical law of sines,

$$ \frac{\sin\frac\pi2}{\sin\alpha} = \frac{\sin\eta}{\sin\beta}.$$

But $\sin\frac\pi2 = 1.$ Therefore $\sin\eta = \frac{\sin\beta}{\sin\alpha}$ and $$\eta = \arcsin\left(\frac{\sin\beta}{\sin\alpha}\right).$$

For now, let's consider the variant of the problem in which $\theta$ is known and $\gamma$ is unknown. Since $\theta$ is known, we can construct two points on the black small circle that subtend angle $\theta$ from $A$. Let the point labeled $B$ in the figure be one of those points.

The spherical triangle $\triangle ABC$ has sides of measure $\alpha$ opposite $A$ and $B$ and a side of measure $\theta$ opposite $C.$ Let $\zeta = \angle BAC.$ By the spherical law of cosines, $$ \cos\alpha = \cos\alpha \cos\theta + \sin\alpha \sin\theta \cos\zeta. $$

Solving for $\zeta,$ $$ \zeta = \arccos\left(\frac{1 - \cos\theta}{\sin\theta}\cot\alpha\right). $$

From $B,$ construct an arc on the sphere to the closest point on the blue great circle. Call that point $E.$ The arc from $B$ to $E$ meets the blue great circle in a right angle.

So we now have a spherical triangle $\triangle ABE$ with a right angle at $E.$ Note that the side $AB$ of this triangle has angular measure $\theta,$ but it is not the arc shown in the diagram. The arc in the diagram is an arc of a small circle, whereas the side of a spherical triangle is always a great circle.

Let $\delta$ be the measure of the arc $BE.$ Let $\phi = \angle BAE.$ Then $\phi = \zeta - \eta,$ where $\zeta$ and $\eta$ have already been computed. Now we have another spherical triangle, this time with a right angle $\angle AEB.$ Applying the spherical law of sines again,

$$ \frac{\sin\frac\pi2}{\sin\theta} = \frac{\sin\eta}{\sin\delta}.$$

Solving for $\delta,$ $$ \delta = \arcsin(\sin\theta \sin\phi). $$

The distance between $B$ and $E$ is also the distance between the two points where the blue circle intersect the $y,z$ plane, namely $D$ and $F$, the point on the blue small circle closest to $D$. Hence $$ \gamma = \beta + \delta. $$

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  • $\begingroup$ Thank your for your time and answer. Is it possible to draw a 3d picture(even by hand is fine)? $\endgroup$
    – T. an
    Mar 15, 2020 at 6:25
  • $\begingroup$ I have adapted the diagram from the question. I also noticed I had written $\alpha$ where I should have written $\beta$ in the final formula, so I fixed that. $\endgroup$
    – David K
    Mar 15, 2020 at 16:14

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