Question

Three distinct points $A,B$ and $C$ are given in the two dimensional coordinate plane such that the ratio of the distance of any one of them from the point $(1,0)$ to the distance from the point $(-1,0)$ is equal to $\frac{1}{3}$. If the circumcentre of the triangle $ABC$ is at the point $(\frac{a}{b},0)a,b\in N$ and coprime, then value of $a+b$ is

Answer 1

Let $P=(1,0),Q=(-1,0),A=(x,y)$
$\frac{AP}{AQ}=\frac{BP}{BQ}=\frac{CP}{CQ}=\frac{1}{3}$
$\Rightarrow 3AP=AQ$
$\Rightarrow 9A{P}^2=A{Q}^2$
$\Rightarrow 9{(x-1)}^2+9y^2={(x+1)}^2+y^2\Rightarrow x^2+y^2-\left(\frac{5}{2}\right)x+1=0\dots \left(1\right)$
Therefore, $A$ lies on a circle.
Similarly, $B$ and $C$ are also lie on the same circle.
Hence, the circum centre of $△ABC$ is centre of circle $\left(1\right)$
which is $(\frac{5}{4},0)$.
$\therefore a+b=5+4=9$