Key Concepts and Formulas

• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• Apollonius' Circle: The locus of a point whose distance from two fixed points is in a constant ratio $k \neq 1$ is a circle.
• Standard Equation of a Circle: The standard form of a circle's equation is $x^2 + y^2 + 2gx + 2fy + c = 0$, where the center of the circle is $(-g, -f)$.

Step-by-Step Solution

Step 1: Setting up the Locus Equation

We are given two points $P(1, 0)$ and $Q(-1, 0)$. Let $A(x, y)$ be any point such that the ratio of its distance from $P$ to its distance from $Q$ is $\frac{1}{3}$. That is, $\frac{AP}{AQ} = \frac{1}{3}$. Our goal is to find the equation of the locus of point A. $\frac{AP}{AQ} = \frac{1}{3}$ Cross-multiplying gives: $3AP = AQ$ Squaring both sides to eliminate the square roots in the distance formula: $(3AP)^2 = (AQ)^2$ $9(AP)^2 = (AQ)^2$ Using the distance formula: $AP^2 = (x - 1)^2 + (y - 0)^2 = (x - 1)^2 + y^2$ $AQ^2 = (x - (-1))^2 + (y - 0)^2 = (x + 1)^2 + y^2$ Substituting these into the equation: $9\left((x-1)^2 + y^2\right) = (x+1)^2 + y^2$

Step 2: Expanding and Simplifying the Equation

Now, we expand the equation and simplify it to get the standard form of a circle equation. $9(x^2 - 2x + 1 + y^2) = x^2 + 2x + 1 + y^2$ Distribute the 9 on the left side: $9x^2 - 18x + 9 + 9y^2 = x^2 + 2x + 1 + y^2$ Move all terms to one side to set the equation to zero: $(9x^2 - x^2) + (9y^2 - y^2) + (-18x - 2x) + (9 - 1) = 0$ $8x^2 + 8y^2 - 20x + 8 = 0$ Divide the entire equation by 8 to get the coefficients of $x^2$ and $y^2$ to be 1: $\frac{8x^2}{8} + \frac{8y^2}{8} - \frac{20x}{8} + \frac{8}{8} = 0$ $x^2 + y^2 - \frac{5}{2}x + 1 = 0 \quad \text{... (Equation 1)}$

Step 3: Identifying the Circumcircle and its Center

Since the three distinct points A, B, and C satisfy the same condition, they all lie on the circle described by Equation 1. This circle is the circumcircle of triangle ABC. The circumcenter is the center of this circle.

Comparing Equation 1 with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$: $x^2 + y^2 - \frac{5}{2}x + 0y + 1 = 0$ We can identify: $2g = -\frac{5}{2} \implies g = -\frac{5}{4}$ $2f = 0 \implies f = 0$ $c = 1$ The center of the circle is $(-g, -f) = \left(-\left(-\frac{5}{4}\right), -0\right) = \left(\frac{5}{4}, 0\right)$. Therefore, the circumcenter of $\triangle ABC$ is $\left(\frac{5}{4}, 0\right)$.

Common Mistakes & Tips

• Algebraic Errors: Carefully expand and simplify the equation to avoid mistakes.
• Center Formula: Remember the correct formula for the center of the circle in standard form is $(-g, -f)$.

Summary

The problem uses the concept of Apollonius' Circle to determine that the three points A, B, and C lie on a circle. This circle is the circumcircle of the triangle ABC. By deriving the equation of the circle and finding its center, we find the circumcenter of the triangle. The circumcenter is $\left(\frac{5}{4}, 0\right)$.

The final answer is $\boxed{\left( {{5 \over 4},0} \right)}$, which corresponds to option (A).