Consider a line in $\mathbb{R}^3$ between $(0,0,2)$ and $(a,b,0)$.
This line intersects $\mathbb{S}_2$ := { $(x,y,z) \in \mathbb{R}^3 | \ x^2 + y^2 + (z-1)^2 = 1 $ } in $(0,0,2)$ and in $X(a,b,0)$.
a) sketch the situation.
b) Compute $X(a,b,0)$. Show that $X^2 : \mathbb{R}^2 \rightarrow \mathbb{R}^3 $ is regular.
c) Compute the first fundamental form.
My work :
So this is how I solved a) : 
So for b) and c) : I know what regular means and I know how to compute the first fundamental form, but my problem is I don't know how to compute $X(a,b)$. And if I haven't $X(a,b)$ I can't show that $X$ is regular and I can't compute the first fundamental form. So my idea is that we can use spherical coordinates? general: $x = \cos(\phi)\sin(\theta), y = \sin(\phi)\sin(\theta), z=\cos(\phi)$. We have a "shifted" unit ball so the spherical coordinates would be: $x = \cos(\phi)\sin(\theta), y = \sin(\phi)\sin(\theta), z=\cos(\phi) +1 $. ( $ \phi \in [0,2\pi) $ and $\theta \in [0,\pi]$ ) . Moreover we know that $X(a,b)$ is on the green triangle ( see the second picture below ) . Can I combine this information to get $X(a,b)$ ? I mean $X(a,b) = ( \cos(\phi)\sin(\theta), \sin(\phi)\sin(\theta),\cos(\phi) +1 $) for a certain $\phi$ and $\theta$. Can we compute $\phi$ in this way? :
$\phi$ = $\cos \measuredangle$ $( (a,0,0) , (a,b,0)) = \frac{<(a,0,0),(a,b,0)>}{||(a,0,0)|| \cdot ||(a,b,0)||} = \frac{a}{\sqrt{(a^2+b^2)}} $ ? Is this right? And if it is : How can I compute $\theta$ ?
