If three distinct points A,B,C are given in the 2-dimensional coordinate plane such that t…

Question

If three distinct points $A,B,C$ are given in the $2$-dimensional coordinate plane such that the ratio of the distance of each one of them from the point $(1,0)$ to the distance from $(-1,0)$ is equal to $	frac12$, then the circumcentre of the triangle $ABC$ is at the point:

Accepted Answer

Let $P=(1,0),\ Q=(-1,0)$. For $A=(x,y)$: $\dfrac{AP}{AQ}=\tfrac12\Rightarrow4AP^2=AQ^2$:
$$4[(x-1)^2+y^2]=(x+1)^2+y^2\Rightarrow3x^2+3y^2-10x+3=0\Rightarrow x^2+y^2-\frac{10}{3}x+1=0.$$

$A,B,C$ all satisfy this same circle, so it is the circumcircle, with centre $\left(\tfrac53,0\right)$.

Solutions · 1