Find the values for $k$ such that the system of equations: $x^2+y^2=k$ and $y^2=4(x-2)$ has no solutions, a unique solution and two solutions? 
Question Find the values for $k$ such that the system of equations: $x^2+y^2=k$ and $y^2=4(x-2)$ has no solutions, a unique solution and two solutions

My attempt:
If we try to solve this problem by substituiton the second equations into the first we get:
$$x^2+4(x-2)=k$$
which has discriminant $48+4k$. Thus the equation has $0,1$ and $2$ solutions when $k<-12$, $k=-12$ and $k>-12$ respectively.
However I notice this doesn't match the picture where it's clear the nature of the solutions changes when $k=4$ , why is this?

The equation $y^2=4(x-2)$ implies that $x\geq2$. That means the curve defined by $y^2=4(x-2)$ (a parabola) has no point for $x<2$. So, if you are looking for its intersection with another curve (here $x^2+y^2=k$) you should look over $x>2$.
For example, if the goal is having no intersection you should find $k$ such that the equation$$x^2+4(x-2)=k$$has no roots over $x>2$. This is equivalent to find $k$ such that the greatest root of the equation happens before $x=2$. Thus by solving the quadratic equation we have $$x^*=\frac{1}{2}\left(-4+\sqrt{48+4k}\right)<2$$ which leads to$$k<4$$
