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Does anybody have formulae to solve the following issue. If you have two circles, defined by their two centres, and a radius for each circle. Where (if the circles intersect) are the two points where the circles intersect (or one if they touch)? I am interested in a solution on a spheroid, not a flat plane.

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This problem is discussed in Section 11 of Geodesics on an ellipsoid of revolution (it's the "trilateration problem").

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Given $C$ is the center of a sphere having radius $r_s$ where we have $r_1,r_2$ as the circle radii along the sphere surface, then they can be used as values in radians by a transformation $r_1\over 2\pi r_s$, and $\rho_1=r_s\sin\left({r_1\over 2\pi r_s}\right)$ is the radius of the circle when flattened. In fact, the sphere having radius $\rho_1$ centered at point $Q_1=C+r_s\cos\left({r_1\over 2\pi r_s}\right)|P_1-C|$ intersects the primary sphere in the required circle. $\rho_2,Q_2$ can be found using the same transformations.

Then we know that $|Q_1-Q_2|\le \rho_1+\rho_2$ is the condition necessary for intersection or tangency of the two spheres. To guarantee intersection or tangency on the sphere surface, one additional condition is necessary which is that the arc length $r_1+r_2$ must be large enough to overcome the distance $|P_1-P_2|$ which will occur when $2r_s\sin\left({r_1+r_2\over 4\pi r_s}\right)\ge |P_1-P_2|$. And once you have that condition satisfied, you can identify $K$ the center of the intersection or tangent point, and use basic trig from there...

In particular, $K$ arises such that

$$r_I^2=\rho_1^2-|Q_1-K|^2=\rho_2^2-|Q_2-K|^2$$

where $r_I$ is the "radius of intersection." The center of intersection on the surface of the sphere will be at the point $L$ which is projected from the center of the main sphere through the point $K$ such that $|L-C|=r_s$. The points of intersection of the two circles will be at the intersection points of the $\rho_1,\rho_2$ sphere intersection circle with the main sphere.

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  • $\begingroup$ Basic trig doesn't work on a spheroid does it? $\endgroup$ – shawn stanley Oct 29 '13 at 14:34
  • $\begingroup$ It does, but the equations are extended. So for a $3$D distance, if you have $(x,y,z)$ as the individual coordinate differences, then the distance is $\sqrt{x^2+y^2+z^2}$ because the $x:y$ hypotenuse has length $d=\sqrt{x^2+y^2}$ and this line is perpendicular to the $z$ line so the $x:y:z$ hypotenuse has length $\sqrt{d^2+z^2}=\sqrt{x^2+y^2+z^2}$. $\endgroup$ – abiessu Oct 29 '13 at 14:38
  • $\begingroup$ Basic trig does not apply on the surface of the spheroid, but that is not the case with your question from what I can tell. $\endgroup$ – abiessu Oct 29 '13 at 14:42
  • $\begingroup$ The original question did state that "I am interested in a solution on a spheroid, not a flat plane". $\endgroup$ – shawn stanley Oct 29 '13 at 16:10
  • $\begingroup$ My apologies, I misread your question to be "the intersection of two spheroids". I will fix my answer to match your question. $\endgroup$ – abiessu Oct 29 '13 at 16:12

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