# Find area of circle from tangent and point on edge

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.

• “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.
– amd
May 17, 2019 at 4:14
• 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. May 17, 2019 at 10:39

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 $$r^2=(p+1)^2+q^2=20$$.
• 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 $B$.