Is there a short and simple way to check if a line is a tangent to a circle, without complicated distance formulae? A solution to a question in my book says that for a circle $(x-at^2)(x-a/t^2) + y(y-2at)=0$ the line x = -a is a tangent. This is only a small part of a much bigger problem I am supposed to do in a few minutes, and I wouldn't like to spend much time on it. a is from parabola $y^2 = 4ax$, where $(at^2,2at)$ is a random point on the parabola. Thus x = -a is directrix.

The exact question and answer as they appear in the book are:

True or false: Directrix of a parabola is the tangent of a circle drawn its focal chord as diameter.

Answer: Equation of such a circle is $(x-at^2)(x-a/t^2) + y(y-2at)=0$. Directrix $x=-a$ which is tangent. Statement is true.

  • 1
    $\begingroup$ This is not clear. $t$ appears to be an unspecified parameter here. Are you asking for which $t$ the given line is tangent to the curve? $\endgroup$ – lulu Apr 15 at 11:01
  • $\begingroup$ @lulu I forgot to mention that a is from parabola y^2 = 4ax, sorry, I'm editing the question now. $\endgroup$ – Hema Apr 15 at 11:56
  • 2
    $\begingroup$ @Hema That edit doesn't change the fact that the statement is false when $a = t = 1,$ as shown in an answer already. Note that you can have $a = 1$ in $y^2=4ax,$ which gives you a parabola $y^2=4x$ with directrix $x=-1,$ and then setting $t=1$ you get the point $(1,2)$ which is a point on the parabola. So the formula is not true for just any random point on the parabola. You might try showing the context for this equation, that is, what did the book say before the equation and just afterwards. $\endgroup$ – David K Apr 15 at 13:19
  • $\begingroup$ @DavidK the exact uestion: "True or false: Directrix of a parabola is the tangent of a circle drawn its focal chord as diameter." Answer: "equation of such a circle is (given euation). Directrix x=-a which is tangent. Statement is true." There is no further info given. $\endgroup$ – Hema Apr 15 at 15:30
  • 1
    $\begingroup$ I took the liberty of copying the book's question into the main question text above. Usually I would ask you to do this yourself to improve the question, but the edit seemed straightforward enough. If I did it wrong, please correct it. $\endgroup$ – David K Apr 15 at 19:58

You have the wrong equation. The other end of the focal chord with one end at $(at^2,2at)$ is $\left(a/t^2,-\frac{2a}t\right)$ and the equation of a circle with that chord as its diameter is

$$ \left(x-at^2\right) \left(x-\frac{a}{t^2}\right) + (y-2at)\left(y + \frac{2a}t\right) =0.$$

The equation as you wrote it in the question (with $y$ instead of $y+\frac{2a}t$) gives you a circle where one end of the diameter has been projected onto the $x$-axis; therefore it is smaller than the correct circle and it will not generally intersect the directrix of the parabola.

Since the directrix is parallel to the $y$-axis, the distance of the center of the circle from the directrix is simply the difference between $-a$ and the $x$-coordinate of the circle's center. You can find this distance, as well as the radius of the circle, from the coordinates of the two endpoints of the focal chord without using the equation of the circle, so it's unclear to me why that equation was brought into play.

It's true that after finding the $y$-coordinate of the center of the circle you can use the equation to determine the $x$-coordinates of the two tangent lines parallel to the $y$-axis, but is that easier than finding the circle's radius and the $x$-coordinate of its center? You could try solving the problem both ways to find out which works better.

Meanwhile there's a solution using synthetic geometry (no coordinates), provided that you recall that the distance from the focus to an arbitrary point $P$ on the parabola is the same as the distance from $P$ to the directrix. In my opinion, that is a much simpler way to solve this.


for a circle $(x-at^2)(x-a/t^2) + y(y-2at)=0$ the line $x = -a$ is a tangent

This is not true, at least not for all $a, t$. For example, take $a=t=1$, and the equation becomes

$$(x-1)^2 + y^2-2y=0$$

which is equivalent to $$(x-1)^2+(y-1)^2 = 1.$$

This is a circle with center $(1,1)$ and radius $1$, which means that the line $x=-a$, which is $x=-1$, is not tangent to it.

  • $\begingroup$ I'm sorry I forgot to mention that a is from parabola y^2 = 4ax, I'm editing the question now. $\endgroup$ – Hema Apr 15 at 11:57

There is actually a very simple way to do that. You need :

--Equation of the circle (or whatever you want if the degrees of the polynomial is $< 3$)

--Equation of the lines $(y = ax + p)$

Simply replace the equation of the line in the equation of the circle (by example replace the "y" from the circle equation with the "$y$" of the line which is "$ax +p$" )

Now you have something like this :

$$ax^{2} + bx + c = 0$$

do : $\Delta = b^{2} - 4ac$

If $\Delta = 0$, then it is tangent. If $\Delta > 0$, it cut the polynome twice If $\Delta < 0$ it never cut the polynome

Note that in this case you will have two different symmetrical equation due to the square in the circle equation. You will have $y = \pm \sqrt{\cdots}$

  • $\begingroup$ For this particular parabola, it’s less work to substitute for $x$. $\endgroup$ – amd Apr 16 at 5:32

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.