If three distinct point A, B, C are given in the 2-dimensional coordinate plane such that the ratio of the distance of each one of them from the point (1, 0) to the distance from (-1, 0) is equal to `(1)/(2)`, then the circumcentre of the triangle ABC is at the point :

Options

A. (3, 0)
B. (5/3, 0)
C. (1/3, 0)
D. (0, 0)

Answer

B

Explanation

Let the coordinates of vertex A be `(alpha , beta)` and let P(1, 0) and Q(-1, 0) be given points such that
`(AP)/(AQ)=(1)/(2)`
`rArr 4 AP^(2)=AQ^(2)`
`rArr 4 {(alpha-1)^(2)+beta^(2)}=(alpha+1)^(2)+beta^(2)`
`rArr alpha^(2)+beta^(2)-(10)/(3) alpha+1=0`
`A(alpha, beta)` lies on `x^(2)+y^(2)-(10)/(3)x+1=0`
Similarly, B and C lie on the circle `x^(2)+y^(2)-(10)/(3)x+1=0`. So, equation of the circumcircle of `Delta ABC` is `x^(2)+y^(2)-(10x)/(3)+1=0`
and hence the coordinates of circumcentre are (5/3, 0).