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

The distance of any point $$P(x,y)$$ on the required locus from $$(1,0)$$ and from $$(-1,0)$$ satisfies

$$\frac{PA}{PB}=\frac{1}{3}, \quad\text{where }A(1,0),\;B(-1,0).$$

Square the ratio so that the algebra involves only polynomials:

$$PA^{2}=\frac{1^{2}}{3^{2}}\,PB^{2}=\frac{1}{9}\,PB^{2} \;-(1)$$

Substitute the distance formulas:

$$\bigl(x-1\bigr)^{2}+y^{2}=\frac{1}{9}\Bigl[(x+1)^{2}+y^{2}\Bigr].$$

Multiply by 9 to clear the denominator:

$$9\bigl[(x-1)^{2}+y^{2}\bigr]=(x+1)^{2}+y^{2}.$$

Expand and collect like terms:

$$9(x^{2}-2x+1)+9y^{2}=x^{2}+2x+1+y^{2}$$ $$\Longrightarrow 8x^{2}-20x+8+8y^{2}=0.$$

Divide by 8 to simplify:

$$x^{2}-\frac{5}{2}x+y^{2}+1=0.$$

Complete the square in $$x$$:

$$(x-\tfrac{5}{4})^{2}-\Bigl(\tfrac{5}{4}\Bigr)^{2}+y^{2}+1=0$$ $$\Longrightarrow (x-\tfrac{5}{4})^{2}+y^{2}=\Bigl(\tfrac{5}{4}\Bigr)^{2}-1 =\frac{25}{16}-\frac{16}{16}=\frac{9}{16}.$$

Hence the locus is a circle with

centre $$\left(\frac{5}{4},0\right)$$ and radius $$\frac{3}{4}.$$

Points $$A,B,C$$ all lie on this same circle, so their circumcircle is exactly this circle. Therefore the circumcentre of $$\triangle ABC$$ is the centre of the above circle, viz.

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

Option B which is: $$\left(\frac{5}{4}, 0\right)$$