Centre of the circle
` (x - x_(1)) (x-x_(2)) + (y-y_(1)) (y- y_(2)) = 0 ` is

Options

A. `((x_(1) + y_(1))/(2) , (x_(1) + y_(2))/(2))`
B. `((x_(1) - y_(1))/(2) , (x_(2) - y_(2))/(2))`
C. `((x_(1) + x_(2))/(2) , (y_(1) + y_(2))/(2))`
D. `((x_(1) - x_(2))/(2) , (y_(1) - y_(2))/(2))`

Answer

To find the center of the circle given by the equation:

\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]

we can follow these steps:

### Step 1: Identify the form of the equation
The given equation represents a circle whose diameter endpoints are \((x_1, y_1)\) and \((x_2, y_2)\).

### Step 2: Understand the concept of the center
The center of a circle is the midpoint of the diameter. The midpoint \(M\) of a line segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) can be calculated using the midpoint formula:

\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]

### Step 3: Apply the midpoint formula
Using the coordinates of the endpoints of the diameter, we can find the center of the circle:

\[ \text{Center} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]

### Conclusion
Thus, the center of the circle is:

\[ \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]