Three distinct points A, B, C are given in a | vedclass

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(D) Let the point be $P(x, y)$. The given condition is $\frac{PA}{PB} = \frac{1}{2}$,where $A = (1, 0)$ and $B = (-1, 0)$.Squaring both sides,we get $

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

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(D) Let the point be $P(x, y)$. The given condition is $\frac{PA}{PB} = \frac{1}{2}$,where $A = (1, 0)$ and $B = (-1, 0)$.Squaring both sides,we get $4PA^2 = PB^2$.$4((x-1)^2 + y^2) = (x+1)^2 + y^2$.$4(x^2 - 2x + 1 + y^2) = x^2 + 2x + 1 + y^2$.$4x^2 - 8x + 4 + 4y^2 = x^2 + 2x + 1 + y^2$.$3x^2 - 10x + 3 + 3y^2 = 0$.$x^2 - \frac{10}{3}x + y^2 + 1 = 0$.This is the equation of a circle with center $C_0 = \left( \frac{5}{3}, 0 \right)$ and radius $r = \sqrt{(\frac{5}{3})^2 - 1} = \sqrt{\frac{25}{9} - 1} = \sqrt{\frac{16}{9}} = \frac{4}{3}$.Since points $A, B, C$ lie on this circle,the circumcenter of triangle $ABC$ is the center of this circle,which is $\left( \frac{5}{3}, 0 \right)$.

Three distinct points $A, B, C$ are given in a two-dimensional coordinate plane such that the ratio of the distance of each point from $(1, 0)$ to its distance from $(-1, 0)$ is equal to $\frac{1}{2}$. What is the circumcenter of triangle $ABC$?

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