The angle between the tangents to the circle with centre (4,5) drawn from P(-2,-3) is `120^(@)` then length of the tagent to the circle from P is

Options

A. a) `5`
B. b) `12`
C. c) `3sqrt(2)`
D. d) `12sqrt(3)`

Answer

To find the length of the tangent from point P(-2, -3) to the circle with center (4, 5) and given that the angle between the tangents is 120 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Center of the circle \( O(4, 5) \)
- Point from which tangents are drawn \( P(-2, -3) \)
- Angle between the tangents \( \angle APB = 120^\circ \)

2. **Determine the Angle at Point O:**
- Since \( \angle APB = 120^\circ \), the angle \( \angle OPA \) (where O is the center of the circle) is half of this angle:
\[ \angle OPA = \frac{180^\circ - 120^\circ}{2} = 30^\circ \]

3. **Calculate the Length of OP:**
- The distance \( OP \) can be calculated using the distance formula:
\[ OP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
- Substituting the coordinates of O and P:
\[ OP = \sqrt{(4 - (-2))^2 + (5 - (-3))^2} \]
\[ = \sqrt{(4 + 2)^2 + (5 + 3)^2} \]
\[ = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]

4. **Use the Cosine of the Angle to Find AP:**
- In triangle \( AOP \), we can use the cosine of angle \( OPA \):
\[ \cos(30^\circ) = \frac{AP}{OP} \]
- We know \( OP = 10 \) and \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ \frac{\sqrt{3}}{2} = \frac{AP}{10} \]
- Solving for \( AP \):
\[ AP = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \]

5. **Conclusion:**
- The length of the tangent from point P to the circle is \( AP = 5\sqrt{3} \).

### Final Answer:
The length of the tangent to the circle from point P is \( 5\sqrt{3} \).