Let's say that we know the circumferences of 3 concentric circles on the surface of a sphere, as well as the distances between the concentric circles and the order of the circles in a given direction, and nothing else. We need to figure out the radius of the sphere given only these 6 values.
In principle this seems like something that would have a unique solution for any set of circumferences and distances between the circles, and without contradictions. In practice I find that actually finding a formula for the surface of the sphere given only these 6 values is challenging.
I know that given a certain radius of a sphere the circumference of a circle is $c=2\pi{R}sin\left(\frac{r}{R}\right)$ with $R$ being the radius of the sphere, $r$ being the radius of a circle along the surface of the sphere, and $c$ being the circumference of the circle on the sphere. I tried to figure out how to solve for $R$ given 3 $c$s and the differences between $r$ but I'm not sure how to.
How do I find the radius of a sphere given the information described?
I hope you'll be able to imagine what I'm describing: Suppose $X,Y,Z$ are the centres of the circles, and $O$ is the centre of the sphere.
Consider a cross-section of the sphere along the direction of the line joining the centres of the circles. This cross-section is a circle. Consider the diameter of this circle which is perpendicular to the other three circles. Then the three smaller radii are perpendicular to the diameter $AB$. Suppose the radius from $X$ meets the CS at $X'$, and so on. Then note that $AX'B$, $AY'B$ and $AZ'B$ are right-angled triangles. Let $XX' = r_1$, $YY' = r_2$, $ZZ' = r_3$, $OA = r$.
Since we are given the order in which the circles lie and their radii, we can figure out which two are on the same side of the centre (may be all three). Suppose $AX>AY>BY>BX$. Now, we note that $$AX = r+\sqrt{r^2-r_1^2}$$and$$AY = r+\sqrt{r^2-r_2^2}$$which can be calculated using similarity. Now, their difference gives the distance $XY$, and the equation can be converted into a quadratic in $r^2$ and solved.