Circle from three points on surface of sphere I need to compute the circle on the surface of a sphere given three points on that very surface.   
It is very easy to do that in Euclidean geometry, but the sphere has no $x$ and $y$, but just two angles and radius. Therefore, I just have the $2 \times 3$ (two angles for each point) angles to achieve that.   
Is there an easy way to do that, or do I have to convert my points to Cartesian and then do the computation? How do I then do the conversion with as much accuracy as possible?
Clarification:
I look for the point on the surface of a sphere to which the 3 points, defined by two angles each, have the same or less angular distance.
 A: To find the angular distance between two points, defined by two angles each, is a classical problem in spherical astronomy and in geography.
If the two coordinates are: $\phi=$right ascension (longitude in geography) and $\theta=$ declination (latitude). Than the angular distance $\alpha$ between two points $A,b$ is given by ( here a proof Great arc distance between two points on a unit sphere, but note that here the angle $\theta$ is that used in spherical coordinates, so it is the complement of the latitude):
$$
\cos \alpha= \sin \theta_A\sin \theta_B+\cos \theta_A\cos \theta_B\cos(\phi_A-\phi_B)
$$
We can use this formula to solve the problem in OP. 
Let $(\phi_C,\theta_C)$ the coordinate of the center of a circle on a spherical surface. The points $P$ of the circle have coordinates $(\phi,\theta)$ such that the angular distance between $P$ and $C$ is a constant value $\rho$ (it is the quotient between the radius of the circle and the radius of the sphere), so the equation of the circe can be written as:
$$
\cos \rho=\sin \theta\sin \theta_C+\cos \theta\cos \theta_C\cos(\phi-\phi_C)
$$
Now we note that this equation contains three parameters $\rho,\phi_C,\theta_C$ so, in principle, we can find this parameters substituting the coordinates of three points for $\phi$ and $\theta$. I never performed these calculations, but they seems a bit complexes.
I suppose that the simpler way to solve the problem in OP is to change to cartesian coordinates and to find the intersection between the plane determined by the three points and the sphere, as suggested in the comments.
