Three distinct points A, B,and C are given in | vedclass

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(D) Let the point be $(h, k)$.Given the ratio of distances is $\frac{1}{3}$:$\frac{\sqrt{(h-1)^2 + k^2}}{\sqrt{(h+1)^2 + k^2}} = \frac{1}{3}$Squaring

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

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(D) Let the point be $(h, k)$.Given the ratio of distances is $\frac{1}{3}$:$\frac{\sqrt{(h-1)^2 + k^2}}{\sqrt{(h+1)^2 + k^2}} = \frac{1}{3}$Squaring both sides:$9((h-1)^2 + k^2) = (h+1)^2 + k^2$$9(h^2 - 2h + 1 + k^2) = h^2 + 2h + 1 + k^2$$9h^2 - 18h + 9 + 9k^2 = h^2 + 2h + 1 + k^2$$8h^2 + 8k^2 - 20h + 8 = 0$Dividing by $8$:$h^2 + k^2 - \frac{5}{2}h + 1 = 0$Completing the square for $h$:$(h^2 - \frac{5}{2}h + \frac{25}{16}) + k^2 = \frac{25}{16} - 1$$(h - \frac{5}{4})^2 + k^2 = \frac{9}{16} = (\frac{3}{4})^2$This is the equation of a circle with center $(\frac{5}{4}, 0)$ and radius $\frac{3}{4}$.Since points $A, B$,and $C$ lie on this circle,the triangle $ABC$ is inscribed in this circle.The circumcenter of any triangle inscribed in a circle is the center of that circle.Therefore,the circumcenter is $(\frac{5}{4}, 0)$.

Three distinct points $A, B$,and $C$ are given in a two-dimensional coordinate plane such that for each point,the ratio of its distance from $(1, 0)$ to its distance from $(-1, 0)$ is equal to $\frac{1}{3}$. What is the circumcenter of triangle $ABC$?

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