I need to find all planes of the form $ax+by+cz=0$ such that the projection of: $$ S=\left\{x^2+y^2+z^2=4\middle|x^2+y^2\leq 2x\right\} $$ Onto that plane is one-to-one, and then use it somehow to find the projected area of this surface to the plane $2x+3y+100z=0$.
I assumed for convenience that $a^2+b^2+c^2=1$ and assumed that the line $(x_0,y_0,z_0)+t(a,b,c)$ intersects the surface on some point with $t\neq 0$. So that point will solve: $$ (x_0+at)^2+(y_0+bt)^2+(c_0+zt)^2=4\Longrightarrow 2(ax_0+by_0+cz_0)t+t^2=1\Longrightarrow t=-2(ax_0+by_0+cz_0) $$ Now I'm trying to find a condition so that this $t$ I found won't solve: $$ (x_0+at)^2+(y_0+bt)^2\leq 1 $$ But since this is an inequality, and since even $x_0^2+y_0^2\leq 2x_0$ (I mean, that point only solves the inequation), I feel like probably hit a dead end.
I want to find some condition on $a,b,c$ so any help would be appreciated.
The forbidden directions are given by the unit vectors $\frac{P-Q}{|P-Q|}$ with $P$ and $Q$ belonging to the surface
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which is made by the points
$$ (x,y,z) = \left(1+\rho\cos\theta,\rho\sin\theta,\pm \sqrt{3-\rho^2-2\rho\cos\theta}\right) $$ for $\theta\in[0,2\pi)$ and $\rho\in[0,1]$. The structure of the difference set $\frac{P-Q}{|P-Q|}$ is not entirely obvious to me, but I guess this could be a good starting point: this set contains all the normalized tangent vectors to the surface.