Tangents drawn from the point `P(1,8)` to the circle `x^2 +y^2 -6x -4y-11=0` touch the circle at the points A&B ifR is the radius of circum circle of triangle PAB then [R]-

Options

A. `x^(2)+y^(2)+4x-6y+19=0`
B. `x^(2)+y^(2)-4x-10y+19=0`
C. `x^(2)+y^(2)-2x+6y-20`
D. `x^(2)+y^(2)-6x-4y+19=0`

Explanation

To solve the problem, we need to find the radius \( R \) of the circumcircle of triangle \( PAB \), where \( P(1, 8) \) is a point from which tangents are drawn to the circle defined by the equation \( x^2 + y^2 - 6x - 4y - 11 = 0 \).

### Step-by-Step Solution:

1. **Rewrite the Circle Equation**:
We start with the circle equation:
\[ x^2 + y^2 - 6x - 4y - 11 = 0 \]
We can rewrite this in standard form by completing the square.

2. **Complete the Square**:
- For \( x \):
\[ x^2 - 6x = (x - 3)^2 - 9 \]
- For \( y \):
\[ y^2 - 4y = (y - 2)^2 - 4 \]
- Substitute back into the equation:
\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 - 11 = 0 \]
\[ (x - 3)^2 + (y - 2)^2 - 24 = 0 \]
\[ (x - 3)^2 + (y - 2)^2 = 24 \]
This shows that the center of the circle \( C \) is \( (3, 2) \) and the radius \( r \) is \( \sqrt{24} = 2\sqrt{6} \).

3. **Find the Length of the Tangents**:
The length of the tangents drawn from point \( P(1, 8) \) to the circle can be calculated using the formula:
\[ L = \sqrt{d^2 - r^2} \]
where \( d \) is the distance from point \( P \) to the center \( C(3, 2) \):
\[ d = \sqrt{(1 - 3)^2 + (8 - 2)^2} = \sqrt{(-2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \]
Now, substituting \( d \) and \( r \):
\[ L = \sqrt{(2\sqrt{10})^2 - (2\sqrt{6})^2} = \sqrt{40 - 24} = \sqrt{16} = 4 \]

4. **Circumradius \( R \) of Triangle \( PAB \)**:
The circumradius \( R \) of triangle \( PAB \) can be calculated using the formula:
\[ R = \frac{abc}{4K} \]
where \( a \), \( b \), and \( c \) are the lengths of the sides of triangle \( PAB \) and \( K \) is the area of the triangle.

Since \( PA \) and \( PB \) are equal (both are tangents), we can denote \( PA = PB = L = 4 \). The distance \( AB \) can be calculated as:
\[ AB = 2L = 2 \times 4 = 8 \]

The area \( K \) of triangle \( PAB \) can be calculated using:
\[ K = \frac{1}{2} \times base \times height \]
Here, the base \( AB = 8 \) and the height is the distance from point \( P \) to line \( AB \). The height can be calculated using the radius \( r \) of the circle:
\[ K = \frac{1}{2} \times 8 \times 2\sqrt{6} = 8\sqrt{6} \]

5. **Final Calculation of \( R \)**:
Now substituting into the circumradius formula:
\[ R = \frac{4 \cdot 4 \cdot 4}{4 \cdot 8\sqrt{6}} = \frac{64}{32\sqrt{6}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3} \]

6. **Final Result**:
The problem asks for \( [R] \), which is the integer part of \( R \):
\[ [R] = 0 \]