Let circles `C_(1), C_(2) and C_(3)` with centres `O_(1), O_(2) and O_(3)` respectively touch each other externally, where
`O_(1)=(-36, 7), O_(2)=(20, 7) and O_(3)=(0, -8)`. The coordinates of the centre of a circle passing through the points of contact of circles `C_(1), C_(2) and C_(2), C_(3) and C_(3),C_(1)` are

Options

A. `(-1,0)`
B. `(1, 0)`
C. `(0, 1)`
D. `(0, -1)`

Answer

To find the coordinates of the center of a circle that passes through the points of contact of the circles \(C_1\), \(C_2\), and \(C_3\), we can use the concept of the incenter of the triangle formed by the centers of the circles. The centers of the circles are given as:

- \(O_1 = (-36, 7)\)
- \(O_2 = (20, 7)\)
- \(O_3 = (0, -8)\)

### Step 1: Calculate the lengths of the sides of the triangle formed by the centers

1. **Calculate the distance \(O_1O_2\)**:
\[ O_1O_2 = \sqrt{(20 - (-36))^2 + (7 - 7)^2} = \sqrt{(20 + 36)^2} = \sqrt{56^2} = 56 \]

2. **Calculate the distance \(O_2O_3\)**:
\[ O_2O_3 = \sqrt{(0 - 20)^2 + (-8 - 7)^2} = \sqrt{(-20)^2 + (-15)^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \]

3. **Calculate the distance \(O_3O_1\)**:
\[ O_3O_1 = \sqrt{(0 - (-36))^2 + (-8 - 7)^2} = \sqrt{(36)^2 + (-15)^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \]

### Step 2: Use the formula for the incenter coordinates

The coordinates of the incenter \(O\) can be calculated using the formula:
\[ O = \left( \frac{A x_1 + B x_2 + C x_3}{A + B + C}, \frac{A y_1 + B y_2 + C y_3}{A + B + C} \right) \]
where:
- \(A = O_2O_3 = 25\)
- \(B = O_3O_1 = 39\)
- \(C = O_1O_2 = 56\)

The coordinates of the centers are:
- \(O_1 = (-36, 7)\)
- \(O_2 = (20, 7)\)
- \(O_3 = (0, -8)\)

### Step 3: Calculate the x-coordinate of the incenter

\[ x = \frac{25 \cdot (-36) + 39 \cdot 20 + 56 \cdot 0}{25 + 39 + 56} \]
Calculating the numerator:
\[ = \frac{-900 + 780 + 0}{120} = \frac{-120}{120} = -1 \]

### Step 4: Calculate the y-coordinate of the incenter

\[ y = \frac{25 \cdot 7 + 39 \cdot 7 + 56 \cdot (-8)}{25 + 39 + 56} \]
Calculating the numerator:
\[ = \frac{175 + 273 - 448}{120} = \frac{0}{120} = 0 \]

### Final Result

Thus, the coordinates of the center of the circle passing through the points of contact of the circles \(C_1\), \(C_2\), and \(C_3\) is:
\[ \boxed{(-1, 0)} \]