4
$\begingroup$

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 )

$\endgroup$
7
  • $\begingroup$ Well, the circle touches at point $(0,0)$ and since the center is at $(2,0)$, then the equation is $(x-2)^2+y^2=4$? $\endgroup$
    – Juniven
    Feb 5 '17 at 14:44
  • $\begingroup$ @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. $\endgroup$
    – Arishta
    Feb 5 '17 at 14:46
  • $\begingroup$ 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. $\endgroup$
    – Juniven
    Feb 5 '17 at 14:49
  • $\begingroup$ 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. $\endgroup$
    – Arishta
    Feb 5 '17 at 14:52
  • $\begingroup$ 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!) $\endgroup$ Feb 5 '17 at 15:16
1
$\begingroup$

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.

$\endgroup$
13
  • $\begingroup$ So you are saying that it is not necessary that eq. (2) will have repeated solution? $\endgroup$
    – Arishta
    Feb 5 '17 at 15:14
  • $\begingroup$ Yes, because you chose the wrong coordinate. $\endgroup$ Feb 5 '17 at 15:16
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 5 '17 at 15:28
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 5 '17 at 15:49
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 5 '17 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.