Find the image of the following circles on the sphere under stereographic projection Find the image of the following circles on the sphere under stereographic projection from the north pole onto the equatorial plane:


*

*$C = S^2 ∩ \{(x_1, x_2, x_3) \;|\; x_1 = x_2\}$

*$C = S^2 ∩ \{(x_1, x_2, x_3) \;|\; x_3 = \frac12\}$

*$C = S^2 ∩ \{(x_1, x_2, x_3) \;|\; x_3 = −\frac12\}$

*$C = S^2 ∩ \{(x_1, x_2, x_3) \;|\; x_1 + x_2 − x_3 = 0\}$


Can anyone do one or two and explain to me?
thank you
 A: My general approach would be as follows: identify the point on the circle closest to the north pole, and the point farthest away from it. Project these two points, e.g. by formulating lines through them and intersecting these with the equatorial plane. Or for $N=(0,0,1)$ you can use the following computation:
\begin{align*}
y_1 &= \frac{x_1}{1-x_3} &
y_2 &= \frac{x_2}{1-x_3} &
y_3 &= 0
\end{align*}
The idea behind this formula is that $1-x_3$ describes the third component of the vector pointing from the north pole to the point $(x_1,x_2,x_3)$ on the sphere. You want to scale things in such a way that this length becomes one, which means you divide by this expression.
The requested circle will pass through these two projected points, and for reasons of symmetry it will be symmetric with respect to the line joining these two points. This is because the original circle on the sphere was symmetric with respect to the great circle through the north pole and these two points. Therefore you can simply take the midpoint between the two projected points as the center for the desired circle.
There are a few special cases to consider. If your spherical circle passes through the north pole, it will be projected to a line, but you can still map the other point, and the desired point will pass through the projection of that second point and will be orthogonal to the line joining that second point to the origin. If your spherical circle has constant distance from the north pole, then its image will be a circle around the origin. Simply map any point to obtain the radius.
I'll leave it at that for now, hope this helps you tackle this problem. I might expand my answer later on.
