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.
Best Answer
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.