Consider $x^{2}+y^{2}-z^{2}= -1$. Now we could make a something like stereographic projection on hyperboloid. Let's call two vertices of hyperboloid $N$(north) and $S$(south). Now for any point on hyperboloid we can easy construct $(x,y,z) \to (u,v)$. And actually there is no problem to show that $\displaystyle x = \frac{2u}{1-u^2-v^2}, y = \frac{2v}{1-u^2-v^2}, z = \frac{1+u^2+v^2}{1-u^2-v^2}$.
Now we want to know the image with projection of intersection with plane $ax+by+cz+d= 0$ and $d\ne 0$. Know if we set $x,y,z$, we will get : $\displaystyle 2au+2bv+c(1+u^2+v^2)+d(1-u^2-v^2) = 0$ and we know that $c \ne 0$, so $(a,b,c,d) \to (\frac{a}{c},\frac{b}{c},1,\frac{d}{c})$, so we have : $2a'u+2b'v+(1+u^2+v^2)+d'(1-u^2-v^2)=0$, so $\displaystyle \left(u+\frac{a'}{1-d'}\right)^{2}+\left(v+\frac{b'}{1-d'}\right)^{2} = \frac{a'^2+b'^{2}+d'^{2}-1}{(1-d')^{2}}$. Now my question is : how can we show that the last one equals to arc of circle (the answer in book tell me so)
I also understand that $|\frac{a}{c}|= |a'| > 1$ or $|\frac{b}{c}|= |b'| > 1$. And if $d' = 1$, then image of projection will be a line. So if we know that $|a'|$ or $|b'|$ more than 1, then RHS will be positive and it will be