Q39 / 90 · JEE Main 2012 (19 May Online) · Mathematics

If three distinct points $A, B, C$ are given in the 2dimensional 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 $\frac{1}{2}$, then the circumcentre of the triangle $A B C$ is at the point

  • Correct answer
    A
    $\left(\frac{5}{3}, 0\right)$
  • B
    $(0,0)$
  • C
    $\left(\frac{1}{3}, 0\right)$
  • D
    $(3,0)$

Solution

Let $P(1,0)$ and $Q(-1,0), A(x, y)$ $$ \begin{aligned} & \text { Given: } \frac{A P}{A Q}=\frac{B P}{B Q}=\frac{C P}{C Q}=\frac{1}{2} \\ & \Rightarrow 2 A P=A Q \\ & \Rightarrow 4(A P)^2=A Q^2 \end{aligned} $$ $$ \begin{aligned} & \Rightarrow 4\left[(x-1)^2+y^2\right]=(x+1)^2+y^2 \\ & \Rightarrow 4\left(x^2+1-2 x\right)+4 y^2=x^2+1+2 x+y^2 \\ & \Rightarrow 3 x^2+3 y^2-8 x-2 x+4-1=0 \\ & \Rightarrow 3 x^2+3 y^2-10 x+3=0 \\ & \Rightarrow x^2+y^2-\frac{10}{3} x+1=0 \end{aligned} $$ $\therefore$ A lies on the circle given by (1). As $B$ and $C$ also follow the same condition. $\therefore \quad$ Centre of circumcircle of $\triangle A B C=$ centre of circle given by (1) $$ =\left(\frac{5}{3}, 0\right) \text {. } $$

Hence, the correct answer is option: 'A'.

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