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Three distinct points A, B and C are given in the 2 - 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 $\frac{1}{3}$. Then the circumcentre of the triangle $A B C$ is at the point
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Verified Answer
The correct answer is:
$\left(\frac{5}{4}, 0\right)$
$\left(\frac{5}{4}, 0\right)$
$$
\begin{aligned}
& P=(1,0) ; Q(-1,0) \\
& \text { Let } A=(x, y) \\
& \frac{A P}{A Q}=\frac{B P}{B Q}=\frac{C P}{C Q}=\frac{1}{3} \\
& \Rightarrow 3 A P=A Q \Rightarrow 9 A P^2=A Q^2 \Rightarrow 9(x-1)^2+9 y^2=(x+1)^2+y^2 \\
& \Rightarrow 9 x^2-18 x+9+9 y^2=x^2+2 x+1+y^2 \Rightarrow 8 x^2-20 x+8 y^2+8=0 \\
& \Rightarrow x^2+y^2-\frac{5}{2} x+1=0
\end{aligned}
$$
$\therefore$ A lies on the circle
Similarly B, C are also lies on the same circle
$\therefore$ Circumcentre of $\mathrm{ABC}=$ Centre of Circle $(1)=\left(\frac{5}{4}, 0\right)$
\begin{aligned}
& P=(1,0) ; Q(-1,0) \\
& \text { Let } A=(x, y) \\
& \frac{A P}{A Q}=\frac{B P}{B Q}=\frac{C P}{C Q}=\frac{1}{3} \\
& \Rightarrow 3 A P=A Q \Rightarrow 9 A P^2=A Q^2 \Rightarrow 9(x-1)^2+9 y^2=(x+1)^2+y^2 \\
& \Rightarrow 9 x^2-18 x+9+9 y^2=x^2+2 x+1+y^2 \Rightarrow 8 x^2-20 x+8 y^2+8=0 \\
& \Rightarrow x^2+y^2-\frac{5}{2} x+1=0
\end{aligned}
$$
$\therefore$ A lies on the circle
Similarly B, C are also lies on the same circle
$\therefore$ Circumcentre of $\mathrm{ABC}=$ Centre of Circle $(1)=\left(\frac{5}{4}, 0\right)$
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