Let circle `C_1 : x^2 + (y-4)^2 = 12` intersects circle `C_2: (x-3)^2 +y^2=13` at A and B. A quadrilateral ACBD is formed by tangents at A and B to both circles. The diameter of circumcircle of quadrilateral ACBD is

Options

A. 4
B. 5
C. 6
D. 9.25

Explanation

To solve the problem, we need to find the diameter of the circumcircle of quadrilateral ACBD formed by the tangents at points A and B where circles \( C_1 \) and \( C_2 \) intersect.

### Step-by-step solution:

1. **Identify the centers and radii of the circles**:
- For circle \( C_1: x^2 + (y-4)^2 = 12 \):
- Center \( O_1 = (0, 4) \)
- Radius \( r_1 = \sqrt{12} = 2\sqrt{3} \)

- For circle \( C_2: (x-3)^2 + y^2 = 13 \):
- Center \( O_2 = (3, 0) \)
- Radius \( r_2 = \sqrt{13} \)

2. **Verify that the circles intersect orthogonally**:
- Two circles intersect orthogonally if the following condition holds:
\[ O_1O_2^2 = r_1^2 + r_2^2 \]
- Calculate \( O_1O_2 \):
\[ O_1O_2 = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
- Now calculate \( r_1^2 + r_2^2 \):
\[ r_1^2 = 12, \quad r_2^2 = 13 \quad \Rightarrow \quad r_1^2 + r_2^2 = 12 + 13 = 25 \]
- Since \( O_1O_2^2 = 25 \), the circles intersect orthogonally.

3. **Determine the coordinates of the centers**:
- The centers of the circles are \( O_1(0, 4) \) and \( O_2(3, 0) \).

4. **Find the length of the diameter of the circumcircle of quadrilateral ACBD**:
- The diameter of the circumcircle can be calculated as the distance between the centers \( O_1 \) and \( O_2 \):
\[ \text{Diameter} = O_1O_2 = 5 \]

### Final Answer:
The diameter of the circumcircle of quadrilateral ACBD is \( 5 \).