If three distinct point 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 `(1)/(2)`, then the circumcentre of the triangle ABC is at the point :

Options

A. (3, 0)
B. (5/3, 0)
C. (1/3, 0)
D. (0, 0)

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Define the points and distances
Let the coordinates of point A be \((H, K)\). The coordinates of points B and C are given as \(B(1, 0)\) and \(C(-1, 0)\). We need to establish the distances from point A to points B and C.

### Step 2: Write the distance formulas
The distance from point A to point B is given by:
\[ AB = \sqrt{(H - 1)^2 + (K - 0)^2} = \sqrt{(H - 1)^2 + K^2} \]
The distance from point A to point C is given by:
\[ AC = \sqrt{(H + 1)^2 + (K - 0)^2} = \sqrt{(H + 1)^2 + K^2} \]

### Step 3: Set up the ratio
According to the problem, the ratio of the distances is given as:
\[ \frac{AB}{AC} = \frac{1}{2} \]
Substituting the distance formulas, we have:
\[ \frac{\sqrt{(H - 1)^2 + K^2}}{\sqrt{(H + 1)^2 + K^2}} = \frac{1}{2} \]

### Step 4: Cross-multiply
Cross-multiplying gives:
\[ 2\sqrt{(H - 1)^2 + K^2} = \sqrt{(H + 1)^2 + K^2} \]

### Step 5: Square both sides
Squaring both sides to eliminate the square roots results in:
\[ 4((H - 1)^2 + K^2) = (H + 1)^2 + K^2 \]

### Step 6: Expand both sides
Expanding both sides:
\[ 4(H^2 - 2H + 1 + K^2) = H^2 + 2H + 1 + K^2 \]
This simplifies to:
\[ 4H^2 - 8H + 4 + 4K^2 = H^2 + 2H + 1 + K^2 \]

### Step 7: Rearrange the equation
Rearranging the equation gives:
\[ 4H^2 - H^2 - 8H - 2H + 4K^2 - K^2 + 4 - 1 = 0 \]
This simplifies to:
\[ 3H^2 - 10H + 3K^2 + 3 = 0 \]

### Step 8: Divide by 3
Dividing the entire equation by 3 yields:
\[ H^2 - \frac{10}{3}H + K^2 + 1 = 0 \]

### Step 9: Identify the circumcircle
This equation represents a circle in the form:
\[ H^2 + K^2 - \frac{10}{3}H + 1 = 0 \]
To find the center of the circumcircle, we can compare it to the standard form of a circle:
\[ X^2 + Y^2 + 2GX + 2FY + C = 0 \]
where the center is \((-G, -F)\).

### Step 10: Find the center
From our equation, we can identify:
- \(G = -\frac{10}{6} = -\frac{5}{3}\)
- \(F = 0\)

Thus, the circumcenter is:
\[ \left(-\left(-\frac{10}{6}\right), 0\right) = \left(\frac{10}{6}, 0\right) = \left(\frac{5}{3}, 0\right) \]

### Final Answer
The circumcenter of triangle ABC is at the point:
\[ \boxed{\left(\frac{5}{3}, 0\right)} \]