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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 )

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    Well, the circle touches at point $(0,0)$ and since the center is at $(2,0)$, then the equation is $(x-2)^2+y^2=4$? – 2017-02-05
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    @ΘΣΦGenSan yeah, this is what my textbook says, but my answer does not match ; my answer contains a constant in the equation of the circle. – 2017-02-05
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    My idea is that, since the circle touches the parabola, with the condition that the center is at the focus, so the radius has to be 2. That's it. – 2017-02-05
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    Yeah, I know this question can be done that way as well, but I am looking for the flaw in my procedure more than I am looking for other possible solutions to this problem. – 2017-02-05
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    I see ambiguous the word "touches". The circle centered at $(2,0)$ of radius 2 is tangent to the parabola at $(0,0)$ and secant at two points. A circle with shorter radius is tangent to the parabola at two points. If the word "touches" means "tangent" there are two solutions (one of them double!) – 2017-02-05
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    @RafaBudría No point on a parabola is nearer to the focus than the vertex. – 2017-02-05
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    @Aretino you ares right. – 2017-02-05

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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.

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    So you are saying that it is not necessary that eq. (2) will have repeated solution? – 2017-02-05
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    Yes, because you chose the wrong coordinate. – 2017-02-05
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    Also, can you explain why will substituting $x=y^2/8$ be different from substituting $y^2=8x$ – 2017-02-05
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    Because the intersection points have two different values of $y$ and those values come to be the same when the curves touch each other. In that case you can use the argument for the resolvent equation to have a single solution. – 2017-02-05
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    No, sorry: I was wrong with my last statement. You can't use the argument with $y$, because the unknown is $y^2$. – 2017-02-05
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    I mean: the two intersection points have different $y$ but the same $y^2$. As the resolvent equation is quadratic in the unknown $y^2$, the argument cannot be applied in this case either. – 2017-02-05
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    I remember being told that whenever two curves intersect with each other, we must get repeated solutions on solving them simultaneously. This made perfect sense to me, because tangency is a special case of secancy. Suppose the parabola in question is fixed and we observe the circle as it first cuts the circle at two points, touches it and finally does not intersect it any distinct points. In the first case, when it touches it, we get two distinct real solutions and when it does not intersect it, we do not get any real solutions. It makes sense to say that when it touches it, we'dget equal root – 2017-02-05
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    Yes, of course what you say is right, but at two conditions: 1. resolvent equation must be quadratic in a certain unknown $X$; 2. intersection points must have different values of $X$, when curves do not touch. In your case condition 2. is not satisfied. – 2017-02-05
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    Okay I get it . So, making the resolvent equation in y must allow me to use the repeated roots logic since it satisfies both the said conditions. I get $y^4+32y^2+64k$=0. Let $y^4=t^2$, then $t^2+32t+64k=0$. On applying D=0 for this quadratic equation , I again get k=4. What is wrong now? – 2017-02-05
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    In this case the unknown is $y^2$, and intersection points have different values of $y$, but the same value for $y^2$. So the method doesn't work in this case either. – 2017-02-05
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    @Aretino, Maybe it's me being confused, but we have a question remaining: what then did Eloise calculate? – 2017-10-08
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    @RafaBudría She got a fake solution, because $k=4$ means $x=-2$ for the tangency point, which is impossible as all points of the parabola have $x\ge0$. – 2017-10-08
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    @Aretino, this is for me, and I can see for others too, a kind of "blind spot". I can see your answer is accurate and true, but I cannot grasp intuitively the flaw in the first reasoning. Thank you anyway. – 2017-10-08