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
$x^2+y^2-4x+k=0\tag1$
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 )
Best Answer
The intersection between parabola and circle consists of two points, having the same $x$. So the argument that the resolvent equation must have a single solution does not work if the unknown is $x$: as a matter of fact, you generally have two solutions for $x$, but one of them must be discarded because negative.
In other words: your approach of making the discriminant of the resolvent quadratic equation to vanish, in order to find tangency, works only if two distinct intersection points have different values of the unknown in the resolvent equation. It works well for a line intersecting a conic, but for two conics it may be ineffective.