We know the Stereographic Projection doesn't preserve areas except for the points on the plane such that $x^2+y^2=1$, because that's where $dA=dxdy$. I was wondering what would happen with lengths: is there a set of points where lengths on the sphere are equal to lengths on the plane?
Stereographic projection and lengths
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In general, points have no length.
But if you mean "are there line segments whose projections have the same length", the answer is "yes, but they're not all that easy to identify and don't seem to have a lot of geometric meaning." (I'm assuming that by a "segment" on a sphere you're willing to consider arcs-of-shortest-distance as segments, and their projections -- which are generally circle-arcs or line segments in the plane -- as the corresponding things whose length we're to measure in the plane.)
Certainly any "segment" on $x^2 + y^2 = 1$ has this property (the arc between the two points, before and after projection, has unchanged length).
A more generally useful property of stereographic projection is the preservation of angles, which holds everywhere.
If "stereographic projection" refers to "projecting the unit sphere centered at the origin away from the north pole $(0, 0, 1)$", then the induced metric on the plane with Cartesian coordinates $(u, v)$ is $$ \left[\frac{2}{1 + u^{2} + v^{2}}\right]^{2} (du^{2} + dv^{2}). $$ Since the term in parentheses is the Euclidean metric, the question of area or length being preserved at a point comes down to asking where the expression in square brackets is equal to unity, namely, along the unit circle $\{u^{2} + v^{2} = 1\}$.
(The geometric reason is clear: The unit sphere intersects the plane of projection along the equator orthogonal to the north pole, and the equator maps to the unit circle in the plane.)