# Find equation of circle with center at focus of parabola $y^2=8x$ which touches the given parabola.

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 )

• Well, the circle touches at point $(0,0)$ and since the center is at $(2,0)$, then the equation is $(x-2)^2+y^2=4$? Feb 5 '17 at 14:44
• @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. Feb 5 '17 at 14:46
• 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. Feb 5 '17 at 14:49
• 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. Feb 5 '17 at 14:52
• 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!) Feb 5 '17 at 15:16

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.
• 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. Feb 5 '17 at 15:28
• 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. Feb 5 '17 at 15:49
• 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. Feb 5 '17 at 16:20