Area of spherical circle under stereographic projection Let $w=\frac1{\sqrt3}(1,-1,1)$. Find the area of the spherical circle C($w, \frac{\pi}6$) under the stereographic projection $\phi(x,y,z)=\left(\frac{x}{1-z}, \frac{y}{1-z}\right)$.
Unfortunately, I have no idea how to do it. Would appreciate some advice.
 A: The image of the circle is an ellipse and to find its area we just need to find its axes. To make things easier, it is convenient to rotate circle $C$ by 45° around $z$-axis, in order to have its axis on the $x-z$ plane: $w={1\over\sqrt3}(\sqrt2,0,1)$. The center of the circle is then $o=w\cos{\pi\over6}={1\over2}(\sqrt2,0,1)$ and its radius is $\sin{\pi\over6}={1\over2}$.
Any point $p$ of the circle can be written as $p=o+r$, where $r$ is any vector perpendicular to $w$ and of length ${1\over2}$. In general, we can write $r$ as $r=(t,s,-\sqrt2t)$, where $s$ and $t$ must satisfy $3t^2+s^2={1\over4}$ in order to have $||r||={1\over2}$. We have then:
$$
p=\left({\sqrt2\over2}+t,\,s,\,{1\over2}-\sqrt2t  \right),
\quad\hbox{with:}\quad
3t^2+s^2={1\over4}.
$$
The stereographic projection of $p$ is then
$$
p'=\left({\sqrt2/2+t\over1/2+\sqrt2t},\,{s\over1/2+\sqrt2t}\right).
$$
To find the endpoints of the major axis of the projected ellipse is easy, because they lie on $x$-axis and correspond to points on the circle with maximum and minimum value of $z$, which is to say with maximum and minimum value of $t$. Plugging $t=\pm{1\over\sqrt{12}}=\pm{1\over2\sqrt3}$ into the formula for $p'$ yields then
$$
x_{\max-\min}=2\sqrt2\pm\sqrt3,
$$
which corresponds to a semi-major axis of $\sqrt3$ and a center of the ellipse located at $(2\sqrt2,0)$.
To find the major axis, observe that its endpoints have the same $x$ as the center of the ellipse, and $x=2\sqrt2$ corresponds to $t=-{1\over3\sqrt2}$ and consequently to $s=\pm{1\over2\sqrt3}$. Plugging these into the formula for $p'$ yields
$$
y_{\max-\min}=\pm\sqrt3.
$$
It turns out that semi-minor axis is $\sqrt3$ too, and our ellipse is just another circle! Its area, of course, is then $3\pi$.
EDIT.
The above answer is more involved than necessary, as it turns out that the image of a circle under stereographic projection is always a circle. It is then enough to find the endpoints of that diameter of the circle on the sphere which is the intersection of the circle with the plane passing through $z$-axis and the center of the circle. The images of those points are the endpoints of a diameter of the image circle.
