MCQ
If three distinct points $A, B, C$ are given in the $2-$ dimensional coordinate plane such that the ratio $(1, 0)$ to the distance from $(-1, 0)$ is equal to $\frac {1}{2},$ then the circumcentre of the triangle $ABC$ is at the point
  • $\left( {\frac{5}{3},0} \right)$
  • B
    $(0,0)$
  • C
    $\left( {\frac{1}{3},0} \right)$
  • D
    $(3,0)$

Answer

Correct option: A.
$\left( {\frac{5}{3},0} \right)$
a
Let $P(1,0)$ and $Q(-1,0)$, $A(x,y)$

Given:$\frac{{AP}}{{AQ}} = \frac{{BP}}{{BQ}} = \frac{{CP}}{{CQ}} = \frac{1}{2}$

$ \Rightarrow 2AP = AQ$

$ \Rightarrow 4{\left( {AP} \right)^2} = A{Q^2}$

$ \Rightarrow 4\left[ {{{\left( {x - 1} \right)}^2} + {y^2}} \right] = {\left( {x + 1} \right)^2} + {y^2}$

$ \Rightarrow 4\left( {{x^2} + 1 - 2x} \right) + 4{y^2} = {x^2} + 12x + {y^2}$

$ \Rightarrow 3{x^2} + 3{y^2} - 8x - 2x + 4 - 1 = 0$

$ \Rightarrow {x^2} + {y^2} - \frac{{10}}{3}x + 1 = 0\,\,\,\,\,\,.....\left( 1 \right)$

$\therefore $ Aline on the circle given by $(1)$ . As $B$ and $C$ also follow the same condition.

$\therefore $Center of circumcirle of 

$\Delta ABC=$ center of circle given by $(1)$

                  $ = \left( {\frac{5}{3},0} \right)$

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