Question

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 $ABC$ is at the point

• A. $(0, 0)$
• B. $\left(\frac{5}{4}, 0\right)$
• C. $\left(\frac{5}{2}, 0\right)$
• D. $\left(\frac{5}{3}, 0\right)$

Answer 1

The Correct Option is B

$P=\left(1, 0\right) ; Q\left(-1, 0\right)$ Let $A=\left(x, y\right)$ $\frac{AP}{AQ}=\frac{BP}{BQ}=\frac{CP}{CQ}=\frac{1}{3}\,...\left(1\right)$ $\Rightarrow 3AP = AQ ? 9AP^{2} = AQ^{2} ? 9\left(x -1\right)^{2} + 9y^{2} = \left(x +1\right)^{2} + y^{2}$ $\Rightarrow 9x^{2} -18x + 9 + 9y^{2} = x^{2} + 2x + 1+ y^{2} ? 8x^{2} - 20x + 8y^{2} + 8 = 0$ $\Rightarrow x^{2}+y^{2}-\frac{5}{2}x+1=0\,...\left(2\right)$ $\therefore$ A lies on the circle Similarly B, C are also lies on the same circle $\therefore$ Circumcentre of $ABC =$ Centre of Circle $\left(1\right)=\left(\frac{5}{4}, 0\right)$