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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 ABC is at the point


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

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