If `A(0, 0), B(1, 0) and C(1/2,sqrt(3)/2)` then the centre of the circle for which the lines `AB,BC, CA` are tangents is

Options

A. `((1)/(2),(1)/(4))`
B. `((3)/(2),(sqrt3)/(2))`
C. `((1)/(2),(1)/(2sqrt3))`
D. `((1)/(2),-(1)/(sqrt3))`

Explanation

To find the center of the circle for which the lines AB, BC, and CA are tangents, we need to determine the incenter of the triangle formed by the points A(0, 0), B(1, 0), and C(1/2, sqrt(3)/2). Since the triangle is equilateral, the incenter can be calculated using the average of the coordinates of the vertices.

### Step-by-Step Solution:

1. **Identify the vertices of the triangle:**
- A = (0, 0)
- B = (1, 0)
- C = (1/2, sqrt(3)/2)

2. **Calculate the coordinates of the incenter:**
The incenter (I) of a triangle can be found using the formula:
\[ I = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
where \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are the coordinates of the vertices A, B, and C respectively.

3. **Substitute the coordinates into the formula:**
\[ I_x = \frac{0 + 1 + \frac{1}{2}}{3} = \frac{1.5}{3} = \frac{1}{2} \]
\[ I_y = \frac{0 + 0 + \frac{\sqrt{3}}{2}}{3} = \frac{\frac{\sqrt{3}}{2}}{3} = \frac{\sqrt{3}}{6} \]

4. **Final coordinates of the incenter:**
Thus, the coordinates of the incenter are:
\[ I = \left( \frac{1}{2}, \frac{\sqrt{3}}{6} \right) \]

5. **Check for the correct option:**
Comparing with the given options, we find that the coordinates \(\left( \frac{1}{2}, \frac{\sqrt{3}}{6} \right)\) correspond to option 3, which is \(\left( \frac{1}{2}, \frac{1}{2}\sqrt{3} \right)\).

### Conclusion:
The center of the circle for which the lines AB, BC, and CA are tangents is \(\left( \frac{1}{2}, \frac{\sqrt{3}}{6} \right)\).