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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.

T. an
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    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. – Eric Towers Jul 24 '19 at 22:36
  • Is there any other name for those? I can edit. – T. an Jul 24 '19 at 22:40
  • 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. – Eric Towers Jul 24 '19 at 23:18
  • @EricTowers Thanks you sir. I am editing according to that. – T. an Jul 24 '19 at 23:22
  • Shouldn't be $AB=2R \sin \theta/2$ ? – Jean Marie Jul 25 '19 at 03:05
  • @JeanMarie Yes sir. My bad. It's a typo. – T. an Jul 25 '19 at 03:05
  • It is very unclear what you mean by "the lower circle's orientation angle from the centre of sphere that respect $\theta$ as it is the rectilinear distance". Usually I would say the two blue circles have the same orientation since their planes are parallel. The angle $\theta$ is not a property of the lower blue circle relative to the sphere; it's a property of the combined positions of the lower blue circle and the black circle. – David K Jul 25 '19 at 11:42
  • You have an unlabeled angle drawn in green where one side seems to be drawn through an arbitrary point inside the disk of the lower blue circle. What is the purpose of drawing that angle? Does it have anything to do with the question? – David K Jul 25 '19 at 11:43
  • @DavidK Sir I have edited the picture for better understanding. I am actually looking for variation from $\beta $ to $\gamma$ in polar angle due to the change of $\theta$ in azimuth angle. – T. an Jul 25 '19 at 15:34
  • OK, I think this actually is different from any of your other questions (though it is hard to keep track) and it is an interesting question for its own sake. – David K Jul 26 '19 at 01:42
  • The angle $\angle AOB$ was labeled $\theta$ in the original diagram. Now it is labeled $\alpha.$ Was that intended? If you want to relate this to some of your other work it seems more convenient to make $\alpha$ be the angle from the $z$ axis to the black circle and to restore the label $\theta$ to the angle $\angle AOB$. – David K Jul 26 '19 at 01:45
  • @DavidK No sir, it was not intended. I have edited again along with your given advice. I actually i got only one partial solved answer for my actual problem and it is not help as my intention is to get and analytical solution for the use of some experimental purpose. – T. an Jul 26 '19 at 17:43
  • I continue to struggle with how to interpret the question. Are the planes of the two blue circles always parallel, or not necessarily parallel? When you write, "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 ...," you make it sound as if the circles are only accidentally parallel in the figure and that the lower circle can be put in a different orientation independently from the blue great circle. – David K Jul 28 '19 at 01:57
  • The other thing that is not completely clear: are $\beta$ and $\gamma$ given, and the problem is to find $\theta$, or is $\theta$ given and the problem is to find $\beta$? If the problem is to find $\beta,$ what is the role of $\gamma$: is it a given angle like $\alpha$, or is it supposed to depend on $\beta$ in some way? – David K Jul 28 '19 at 02:10
  • @DavidK Sir both blue circles are always parallel. As $\beta$ define the orientation of great circle. In this problem $\beta $ and $\theta$ is known. But $\gamma $ need to define in a such way that if I change my $\theta$ then$\gamma$ should change accordingly to keep both circle parallel to each other. On the other hand if I choose $\gamma$ to be a known value then $\theta$ should be define in such a way that it will keep both circle parallel to each other. – T. an Jul 28 '19 at 04:38

2 Answers2

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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 $k = a^2 b^2 - 2a^2 - b^2 + 2a^2 \cos \theta - \cos^2\theta$

Peter Taylor
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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 $$ \gamma = \beta + \delta. $$

David K
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  • Thank your for your time and answer. Is it possible to draw a 3d picture(even by hand is fine)? – T. an Mar 15 '20 at 06:25
  • 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. – David K Mar 15 '20 at 16:14