# 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 Option 1) $\left ( \frac{5}{4},0 \right )$ Option 2) $\left ( \frac{5}{2},0 \right )$ Option 3) $\left ( \frac{5}{3},0 \right )$ Option 4) $\left ( 0,0 \right )$

As we learnt in

Circumcentre -

The point of intersection of the perpendicular bisectors of the sides of a triangle.

- wherein

The centre of the circumcircle of a triangle.

Let x be (h,k)

A,B, C lie on locus of x.

given that $\frac{PA}{PB}=\frac{1}{3}$

$\Rightarrow 9PA^{2}=9PB^{2}$

$\Rightarrow 9\left(\sqrt{(h^{2}-1^{2})+k^{2}} \right )^{2}=(h+1)^{2}+k^{2}$

$\Rightarrow 8h^{2}+8k^{2}-20h+8=0$

$\Rightarrow h^{2}+k^{2}-\frac{5}{2}h+1=0$

Centre is $(\frac{5}{4},0)$

Circumstance is $(\frac{5}{4},0)$

Option 1)

$\left ( \frac{5}{4},0 \right )$

This option is correct.

Option 2)

$\left ( \frac{5}{2},0 \right )$

This option is incorrect.

Option 3)

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

This option is incorrect.

Option 4)

$\left ( 0,0 \right )$

This option is incorrect.

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