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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 `1:3.` Then the circumcentre of the triangle `ABC` is at the point (1) `(0,0)` (2) `(5/4,0)` (c) `(5/2,0)` (d) `(5/3,0)`

Text Solution

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P(1,0) Q(-1,0)
`(AP)/(AQ)=1/3,(BP)/(BQ)=1/3,(CP)/(CQ)=1/3`
A(x,y)
`d=sqrt((x_1-x_2)^2+(y_1-y^2)^2+(z_1-z_2)^2)`
`d^2=(x_1-x_2)^2+(y_1-y_2)^2`
`(AP^2)/(AQ^2)=1/9`
`((x-1)^2+y^2)/((x+1)^2+y^2)=1/9`
`9x^2-18x+9+9y^2=x^2+2x+1+y^2`
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