If three distinct points $$A, B, C$$ 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 $$\frac{1}{2}$$, then the circumcentre of the triangle $$ABC$$ is at the point

Let $$P(x,y)$$ be any point that satisfies the given condition

$$\frac{\text{distance of }P\text{ from }(1,0)}{\text{distance of }P\text{ from }(-1,0)}=\frac12$$

Using the distance formula, this gives

$$\frac{\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}}=\frac12$$

Square both sides:

$$(x-1)^2+y^2=\frac14\bigl[(x+1)^2+y^2\bigr]$$

Multiply by $$4$$ to clear the fraction:

$$4\bigl[(x-1)^2+y^2\bigr]=(x+1)^2+y^2$$

Expand the brackets:

$$4\bigl[x^2-2x+1+y^2\bigr]=x^2+2x+1+y^2$$

$$4x^2-8x+4+4y^2=x^2+2x+1+y^2$$

Bring all terms to the left side:

$$3x^2-10x+3y^2+3=0$$

Divide by $$3$$ to get the standard quadratic form:

$$x^2- \frac{10}{3}x+y^2+1=0$$

Rewrite as $$x^2+y^2+\frac{10}{3}(-x)+1=0$$ and complete the square in $$x$$:

$$\bigl[x^2-\frac{10}{3}x+\bigl(\frac{5}{3}\bigr)^2\bigr]+y^2+1-\bigl(\frac{5}{3}\bigr)^2=0$$

$$\bigl[x-\frac{5}{3}\bigr]^2+y^2=\bigl(\frac{5}{3}\bigr)^2-1$$

Simplify the right-hand side:

$$\bigl[x-\frac{5}{3}\bigr]^2+y^2=\frac{25}{9}-\frac{9}{9}=\frac{16}{9}=\bigl(\frac{4}{3}\bigr)^2$$

Thus every point satisfying the ratio condition lies on the circle

$$\bigl[x-\frac{5}{3}\bigr]^2+y^2=\bigl(\frac{4}{3}\bigr)^2$$

This is the Apollonius circle for the two foci $$(1,0)$$ and $$(-1,0)$$ with ratio $$\frac12$$. Its centre is

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

Since the three given points $$A,B,C$$ all lie on this circle, their circumcircle is exactly this Apollonius circle. Therefore the circumcentre of $$\triangle ABC$$ is the centre of this circle, i.e.

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

Option A which is: $$\left(\frac{5}{3},0\right)$$