Three distinct points A, B and C are given in the 2aedimensional 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 `1/3`.Then the circumcentre of the triangle ABC is at the point :

Options

A. `(0,0)`
B. `((5)/(4),0)`
C. `((5)/(2),0)`
D. `((5)/(3),0)`

Explanation

To solve the problem, we need to find the circumcenter of triangle ABC given that the ratio of the distance from any point A, B, or C to the point (1, 0) and the point (-1, 0) is 1/3. Let's denote the points as follows:

- Let \( P = (1, 0) \)
- Let \( Q = (-1, 0) \)
- Let \( A = (x, y) \)

### Step 1: Set up the distance ratio
According to the problem, we have:
\[ \frac{AP}{AQ} = \frac{1}{3} \]
This implies:
\[ AP = \frac{1}{3} AQ \]

### Step 2: Use the distance formula
Using the distance formula, we can express \( AP \) and \( AQ \):
\[ AP = \sqrt{(x - 1)^2 + (y - 0)^2} = \sqrt{(x - 1)^2 + y^2} \]
\[ AQ = \sqrt{(x + 1)^2 + (y - 0)^2} = \sqrt{(x + 1)^2 + y^2} \]

### Step 3: Square both sides to eliminate the square roots
Squaring both sides of the distance ratio gives:
\[ AP^2 = \left(\frac{1}{3}\right)^2 AQ^2 \]
Substituting the expressions for \( AP \) and \( AQ \):
\[ (x - 1)^2 + y^2 = \frac{1}{9} \left((x + 1)^2 + y^2\right) \]

### Step 4: Clear the fraction
Multiplying both sides by 9 to eliminate the fraction:
\[ 9((x - 1)^2 + y^2) = (x + 1)^2 + y^2 \]

### Step 5: Expand both sides
Expanding both sides:
\[ 9(x^2 - 2x + 1 + y^2) = x^2 + 2x + 1 + y^2 \]
This simplifies to:
\[ 9x^2 - 18x + 9 + 9y^2 = x^2 + 2x + 1 + y^2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 9x^2 - x^2 + 9y^2 - y^2 - 18x - 2x + 9 - 1 = 0 \]
\[ 8x^2 + 8y^2 - 20x + 8 = 0 \]
Dividing the entire equation by 4:
\[ 2x^2 + 2y^2 - 5x + 2 = 0 \]

### Step 7: Completing the square
To find the circumcenter, we need to complete the square for the \( x \) terms:
\[ 2(x^2 - \frac{5}{2}x) + 2y^2 + 2 = 0 \]
Completing the square:
\[ 2\left(x^2 - \frac{5}{2}x + \left(\frac{5}{4}\right)^2 - \left(\frac{5}{4}\right)^2\right) + 2y^2 + 2 = 0 \]
\[ 2\left((x - \frac{5}{4})^2 - \frac{25}{16}\right) + 2y^2 + 2 = 0 \]
\[ 2(x - \frac{5}{4})^2 + 2y^2 - \frac{25}{8} + 2 = 0 \]
\[ 2(x - \frac{5}{4})^2 + 2y^2 = \frac{9}{8} \]

### Step 8: Identify the circumcenter
The circumcenter of triangle ABC is the center of the circle represented by the equation:
\[ (x - \frac{5}{4})^2 + y^2 = \frac{9}{16} \]
Thus, the circumcenter is:
\[ \left(\frac{5}{4}, 0\right) \]

### Final Answer
The circumcenter of triangle ABC is at the point \( \left(\frac{5}{4}, 0\right) \).