Find equation of circle with center at focus of parabola $y^2=8x$ which touches the given parabola.
My attempt : Focus of the given parabola is $(2,0)$
Therefore, equation of required circle is
On solving the parabola and the above circle simultaneously, we must get the point of tangency. On substituting $y^2=8x$ in $(1)$ ,
we get $x^2+4x+k=0\tag2$
Now $(2)$ must be a perfect square since the circle touches the parabola. If this equation had $2$ distinct roots, then it would mean that the parabola intersects the circle at two distinct points. For $(2)$ to be a perfect square, $k=4$
Therefore, the required equation of circle is $x^2+y^2-4x+4=0$
The answer given in my textbook is $x^2+y^2-4x=0$. Also, if $k=4$ as I obtained above then $r^2=-8$ since $k= -(r^2+4)$ .
Where am I wrong? ( I am not looking for more possible solutions to this questions )