2
$\begingroup$

I'm helping someone to prepare for a math exam and one of the example questions was as follows:

A circle goes through point $A(-1,0)$ and is tangent to the line with equation $y = 2x$ in the point $B(1,2)$. What is the area of the circle?

I solved it, but in a very messy way. Let the circle have center $(p,q)$ and radius $r$

The circle being tangent to $y = 2x$ means the perpendicular to this line in the point $B$ will pass through the center of the circle. The equation of this line is $y = -\frac{1}{2}(x-1) +2$.

Knowing this and that the circle goes through $A$ and $B$ gives us three equations in total $$ (-1 - p)^2 + (0 - q)^2 = r^2 $$ $$ (1 - p)^2 + (2 - q)^2 = r^2 $$ $$ -\frac{1}{2}p + \frac{5}{2} = q$$

With a ton of algebra it's possible to solve this system, giving the solution $r^2 = 20$ and thus an area of $20\pi$. This seemed like a lot of work for an exam aimed at high school graduates, so I'm wondering if there is a more simple and geometric solution.

$\endgroup$
2
  • $\begingroup$ “A ton of algebra?” Subtract the second equation from the first to eliminate all of the squared terms. You end up with a system of two linear equations in two unknowns. Not coincidentally, the line that the difference of the first two equations represents is exactly the bisector mentioned in this answer below. $\endgroup$
    – amd
    May 17, 2019 at 4:14
  • $\begingroup$ Yes @amd, a ton of algebra given that you only have a few minutes for every answer. For a random high school student that's pretty tight. This question is an outlier compared to the others of the exam, which is why I wondered if I missed an obvious easy answer. But I don't have a problem accepting that this is the easiest way, and that some questions on the exam are just harder than others. $\endgroup$
    – Kasper
    May 17, 2019 at 10:39

1 Answer 1

5
$\begingroup$

You can avoid the equation of the circle by using $(p,q)$ lies on the perpendicular bisector of $AB$ (since the circle passing through both $A$ and $B$), giving an equation $p+q=1$. Solving $$ \left\{ \begin{aligned} p+q&=1\\ p+2q&=5 \end{aligned} \right. $$ gives $(p,q)=(-3,4)$. So $r^2=(p+1)^2+q^2=20$.

$\endgroup$
4
  • $\begingroup$ This bisector (a.k.a. the circles’ radical axis) is exactly the line that corresponds to the equation obtained by computing the difference of the first two equations in the system developed in the question. $\endgroup$
    – amd
    May 17, 2019 at 4:17
  • $\begingroup$ @amd there is only one circle here, but yes if you consider circles centred A,B instead (i.e. the circles defined by the first two equations) that would be their radical axis. $\endgroup$ May 17, 2019 at 4:43
  • $\begingroup$ Per the first two equations in the system, the center of the circle to be found lies on one of the intersections of equal-radius circles centered at $A$ and $B$. $\endgroup$
    – amd
    May 17, 2019 at 4:45
  • $\begingroup$ I guess that's as easy as it'll get. Neat trick $\endgroup$
    – Kasper
    May 17, 2019 at 10:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .