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 circles \( C_1, C_2 \) and \( C_2, C_3 \) and \( C_3, C_1 \), we need to determine the incenter of triangle formed by the centers of these circles \( O_1, O_2, O_3 \).

### Step 1: Identify the coordinates of the centers
The centers of the circles are given as:
- \( O_1 = (-36, 7) \)
- \( O_2 = (20, 7) \)
- \( O_3 = (0, -8) \)

### Step 2: Calculate the lengths of the sides of triangle \( O_1O_2O_3 \)
Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):

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

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

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

### Step 3: Use the lengths to find the incenter
The incenter \( I \) of triangle \( O_1O_2O_3 \) can be calculated using the formula:
\[ I_x = \frac{a \cdot x_1 + b \cdot x_2 + c \cdot x_3}{a + b + c} \]
\[ I_y = \frac{a \cdot y_1 + b \cdot y_2 + c \cdot y_3}{a + b + c} \]
where \( a, b, c \) are the lengths of the sides opposite to vertices \( O_1, O_2, O_3 \) respectively.

Assigning:
- \( a = O_2O_3 = 25 \)
- \( b = O_3O_1 = 39 \)
- \( c = O_1O_2 = 56 \)

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

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

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

### Step 4: Conclusion
The coordinates of the center of the circle passing through the points of contact of circles \( C_1, C_2 \), \( C_2, C_3 \), and \( C_3, C_1 \) is:
\[ \boxed{(-1, 0)} \]