# Image of intersection of a plane with S$^{2}$ under stereographic projection

I have the plane H and its equation is x+y-z = 0. I want to find the image of H$\cap\mathbb{S}^{2}$ under stereographic projection.

I basically said that if the point (x,y,z) gets mapped to say (p,q,0) then the pre-image of (p,q,0), which is ($\frac{2p}{p^{2}+q^{2}+1}$,$\frac{2q}{p^{2}+q^{2}+1}$,$\frac{p^{2}+q^{2}-1}{p^{2}+q^{2}+1}$), must satisfy the equation of the plane.

That gives us 2p+2q = p$^{2}$+q$^{2}$-1

$\Rightarrow$ (p-1)$^{2}$ + (q-1)$^{2}$ = 3 , which is equation of a circle with center at (1,1) and radius $\sqrt{3}$.

Any comment on my work would be appreciated.

• It seems flawless. – user228113 Jun 13 '17 at 19:43
• Yes this seems good ! – user171326 Jun 13 '17 at 19:44
• Thank you very much! I just started self learning Complex Analysis from free textbook on Matthias Beck's San Francisco State University's website like a week ago. So I try asking questions to check.. – HumbleStudent Jun 14 '17 at 2:08
• HumbleStudent, what did you do with the unit normal vector? – BCLC Aug 9 '18 at 7:30

A way to check your answer is that the image of the intersection of $x+y=z, x^2+y^2+z^2=1$ is $(\frac{x}{1-z},\frac{y}{1-z},0) = (\frac{x}{1-(x+y)},\frac{y}{1-(x+y)},0)$ and should be still $C[(1,1),\sqrt{3}]$. I think it's kinda hard to see that $(\frac{x}{1-(x+y)},\frac{y}{1-(x+y)},0)$ is a circle in $\mathbb C$ (well, circle, cline, ellipse, whatever it is, it's obviously in $\mathbb C$) w/c is why using the inverse formula (which I did too), is better I guess. Anyhoo, we can check that
$$(\frac{x}{1-(x+y)}-1)^2+(\frac{y}{1-(x+y)}-1)^2=3,$$