The angle between a pair of tangents from a point P to the circle `x^(2)+y^(2)-6x-8y+9=0` is `(pi)/(3)`. Find the equation of the locus of the point P.

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem, we need to find the equation of the locus of the point \( P \) from which the angle between the tangents drawn to the circle is \( \frac{\pi}{3} \).

### Step-by-step Solution:

1. **Identify the Circle's Equation**:
The equation of the circle is given as:
\[ x^2 + y^2 - 6x - 8y + 9 = 0 \]

2. **Convert the Circle's Equation to Standard Form**:
We will complete the square for \( x \) and \( y \).

- For \( x \):
\[ x^2 - 6x \rightarrow (x - 3)^2 - 9 \]
- For \( y \):
\[ y^2 - 8y \rightarrow (y - 4)^2 - 16 \]

Substituting these back into the circle's equation, we get:
\[ (x - 3)^2 - 9 + (y - 4)^2 - 16 + 9 = 0 \]
Simplifying this gives:
\[ (x - 3)^2 + (y - 4)^2 - 16 = 0 \]
Thus, the equation of the circle is:
\[ (x - 3)^2 + (y - 4)^2 = 16 \]
This indicates that the center of the circle is \( (3, 4) \) and the radius \( r = 4 \).

3. **Using the Angle Between Tangents**:
The angle \( \theta \) between the tangents from point \( P \) to the circle is given as \( \frac{\pi}{3} \). The formula relating the angle \( \theta \), the distance \( OP \) from the center \( O \) to the point \( P \), and the radius \( r \) of the circle is:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{r}{d} \]
where \( d = OP \).

For \( \theta = \frac{\pi}{3} \):
\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} = \frac{4}{d} \]
Thus, we can solve for \( d \):
\[ d = 4 \sqrt{3} \]

4. **Finding the Locus of Point \( P \)**:
The locus of point \( P \) will be a circle with center \( O(3, 4) \) and radius \( OP = 4\sqrt{3} \).

Therefore, the equation of the locus is:
\[ (x - 3)^2 + (y - 4)^2 = (4\sqrt{3})^2 \]
Simplifying gives:
\[ (x - 3)^2 + (y - 4)^2 = 48 \]

5. **Final Equation of the Locus**:
The final equation of the locus of point \( P \) is:
\[ (x - 3)^2 + (y - 4)^2 = 48 \]