How can I find the set of the centers of the circles on sphere that pass through 2 given points and have pretedermined radius, using spherical coordinates? Assume that the radius of sphere is $R$.
(Thanks for helping me to point out the circle is not unique. I think there are 2 such circles.)
Thank you for your help.
On
Characterizing the circle as intersection with a plane:
Assuming the center of the sphere is at the origin.
All points of a circle lie within a plane.
If we intersect a plane with a sphere then for points in the intersection these equations hold: $$ (n_x, n_y, n_z) \cdot (x, y, z) = d \\ x^2 + y^2 + z^2 = R^2 $$ where $n = (n_x, n_y, n_z)^T$ is a normal unit vector of the plane and $d$ is the distance of the plane to the origin.
We can move both objects together (their relative position to each other is kept) around the origin such that $$ n = (0, 0, 1)^T $$ ($n$ starting on the plane is always parallel to $n'$ starting at the origin, by rotating the sphere around its origin we can orient $n'$ to point upwards and thus $n$ as well)
This changes the equations to: $$ z = d \\ x^2 + y^2 + z^2 = R^2 $$ and together give $$ x^2 + y^2 = R^2 - d^2 $$ So the intersection is non-empty for $0 < \lvert d \rvert \le R$. Its points form a circle having radius $r = \sqrt{R^2 - d^2}$ and origin $u = d\,n = (0,0,d)^T$.
Specifying the circle:
If we specify the plane, we have specified the circle.
Two distinct points $P$ and $Q$ on the circle, thus two distinct points on the plane, are not enough to fix that plane.
One could still rotate the plane around the line through both given points, rotating the circle with it, the point staying fixed, and on both old and new circle.
So what about the given radius?
The above image shows the case of the rotation axis not going through the origin. So we have different radii with different rotation angle. However the radius seems to repeat for a rotation angle of 90 degrees. So this description is not unique.
If the rotation axis goes through the origin, all resulting circles have the same radius $R$.
So what we need is three different points on the circle.
Or a unit normal vector to the circle, which is the unit normal of the plane, and the distance of the plane from the origin.
Call the two given points $P,Q$ and let $r$ be the radius of the circle. Moreover we may assume that the sphere is centred at $(0,0,0)$ and that $P = (0,0,R)$, and let's write the coordinates of $Q$ as $(q_x,q_y,q_z)$. We will denote by $M$ the radius of a great circle and by $d(A,B)$ the great circle distance between any two points $A,B$ on the sphere.
Now, clearly if $r > M$ or if $r < \frac{1}{2} d(P,Q)$ there can be no solution. On the other hand, if $r = \frac{1}{2} d(P,Q)$ there is exactly one circle through $P$ and $Q$, with centre at the midpoint between them.
Otherwise, observe that a circle is just the locus comprised of all points a fixed distance from a given point, therefore a solution is given by any of the points of intersection of the circles of radius $r$ and centre $P$ and $Q$, respectively. Note that this gives exactly two solutions unless $P$ and $Q$ are antipodal, in which case we must have $r = M$ and there are infinitely many solutions, whose locus is the great circle cut by the plane $z = 0$.
Instead of actually computing the equations of those two circles, though, it is easier to write down the equations of the planes $\pi_P$ and $\pi_Q$ that cut them. Then we just have to compute the intersection between those two planes and the sphere.
In order to do this, though, I am going to assume that we know how to compute the coordinates of a point $(x_P,y_P,z_P)$ on the sphere at distance $r$ from $P$, and those of a point $(x_Q,y_Q,z_Q)$ on the sphere at distance $r$ from $Q$. Why? Because the plane that cuts a circle with centre a given point on the sphere is orthogonal to the line through that point and the origin. This means that $$ \begin{gather} \pi_P \colon \;\; z = z_Q\\ \text{and}\\ \pi_Q \colon \;\; q_x x + q_y y + q_z z = q_x x_Q + q_y y_Q + q_z z_Q =: c_Q. \end{gather} $$ Therefore the coordinates of the centres to the circles through $P$ and $Q$ of radius $r$ are given by the system $$ \begin{cases} z = z_Q\\ q_x x + q_y y + q_z z = c_Q \\ x^2 + y^2 + z^2 = R^2 \end{cases} $$ and solving it boils down to solving a quadratic equation in one variable.