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 `1/3`.Then the circumcentre of the triangle ABC is at the point :

Options

A. (0, 0)
B. (5/4, 0)
C. (5/2, 0)
D. (5/3, 0)

Answer

B

Explanation

Let (a, y) be the coordinates of point A and P(1, 0) and Q(-1, 0) be given points such that
`(AP)/(AQ)=(BP)/(BQ)=(CP)/(CQ)=(1)/(3)`
`rArr ` A, B, C lie on a circle
`rArr` Circumcentre of `DeltaABC` is the centre of the circle passing through A, B and C.
In order to find the equation of the circle, we proceed as follows:
We have, `(AP)/(AQ)=(1)/(3)`
`rArr 3AP=AQ`
`rArr 9 AP^(2)=AQ^(2)rArr 9 {(x-1)^(2)+y^(2)}=(x+1)^(2)+y^(2)`
`rArr 8x^(2)+8y^(2)-20x+8=0 rArr 2 (x^(2)+y^(2))-5x+2=0`
Hence, the coordinates of the circumcentre are (5/4, 0).