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Step-by-Step Solution
Step 1: Understand the Given Condition
We have three points A, B, and C in the plane, each of which satisfies the ratio condition:
$ \frac{AP}{AQ} = \frac{BP}{BQ} = \frac{CP}{CQ} = \frac{1}{3} $
where $P(1,0)$ and $Q(-1,0)$.
Step 2: Convert the Ratio to an Equation
The ratio $\frac{AP}{AQ} = \frac{1}{3}$ implies $3 \, AP = AQ$. To work with squared distances (which are easier to handle), we square both sides:
$ 3 \, AP = AQ \quad \Longrightarrow \quad 9 \, AP^2 = AQ^2.
$
Step 3: Express $AP^2$ and $AQ^2$ in Terms of Coordinates
Let $A = (x,y)$. Then
$ AP^2 = (x - 1)^2 + y^2 $ (distance from $A$ to $P(1,0)$),
$ AQ^2 = (x + 1)^2 + y^2 $ (distance from $A$ to $Q(-1,0)$).
Thus, the condition $9\,AP^2 = AQ^2$ becomes:
$ 9 \bigl[(x - 1)^2 + y^2 \bigr] = (x + 1)^2 + y^2.
$
Step 4: Simplify the Resulting Equation
Expand each side:
$ 9\bigl(x^2 - 2x + 1 + y^2 \bigr) = x^2 + 2x + 1 + y^2.
$
$ 9x^2 - 18x + 9 + 9y^2 = x^2 + 2x + 1 + y^2.
$
Combine like terms:
$ 8x^2 - 20x + 8y^2 + 8 = 0.
$
Divide by 8 if desired, or rearrange to get:
$ x^2 + y^2 - \frac{5}{3}x + 1 = 0.
$
Step 5: Observe That A, B, and C Lie on the Same Circle
Because B and C must satisfy the same ratio condition, B and C also lie on the same circle given by:
$ x^2 + y^2 - \frac{5}{3}x + 1 = 0. $
This circle contains all points whose distance ratio from $(1,0)$ and $(-1,0)$ is $\frac{1}{3}.$
Step 6: Identify the Center of This Circle (and Hence the Circumcenter)
A circle in the form
$ x^2 + y^2 - 2ax - 2by + c = 0 $
has its center at $(a,b)$. Comparing that with
$ x^2 + y^2 - \frac{5}{3}x + 1 = 0, $
we see that only the $x$-coordinate is shifted by $\frac{5}{6}$ in the standard factor of 2. More precisely, rearranging we see the center is at:
$ \left(\frac{5}{4}, 0\right). $
Therefore, the circumcenter of triangle $ABC$ is
$ \left(\frac{5}{4},0\right). $
