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 and B. The equation of the circumcircle of the triangle PAB is

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-29=0`
D. `x^(2)+y^(2)-6x-4y+19=0`

Answer

To solve the problem, we need to find the equation of the circumcircle of triangle PAB, where P is the point (1, 8) and A and B are the points where the tangents from P touch the given circle.

### Step-by-Step Solution:

1. **Identify the Circle Equation**:
The given circle is represented by the equation:
\[ x^2 + y^2 - 6x - 4y - 11 = 0 \]

2. **Rewrite the Circle Equation in Standard Form**:
We can complete the square for both x and y to find the center and radius of the circle.
\[ (x^2 - 6x) + (y^2 - 4y) = 11 \]
Completing the square:
\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 = 11 \]
\[ (x - 3)^2 + (y - 2)^2 = 24 \]
Thus, the center of the circle \( C \) is \( (3, 2) \) and the radius \( r = \sqrt{24} = 2\sqrt{6} \).

3. **Find the Length of the Tangents from Point P to the Circle**:
The length of the tangents from point \( P(1, 8) \) to the circle can be calculated using the formula:
\[ \text{Length of tangent} = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]
where \( (h, k) \) is the center of the circle and \( r \) is the radius.
\[ = \sqrt{(1 - 3)^2 + (8 - 2)^2 - (2\sqrt{6})^2} \]
\[ = \sqrt{(-2)^2 + (6)^2 - 24} \]
\[ = \sqrt{4 + 36 - 24} = \sqrt{16} = 4 \]

4. **Find the Coordinates of Points A and B**:
The points A and B where the tangents touch the circle can be found using the fact that they lie on the line connecting P and C, and are at a distance of 4 from P along the direction of the tangents. However, we don't need the exact coordinates of A and B to find the circumcircle.

5. **Find the Equation of the Circumcircle of Triangle PAB**:
The circumcircle of triangle PAB will have its center at the midpoint of line segment PC, where P is (1, 8) and C is (3, 2).
Midpoint \( M \) of \( P(1, 8) \) and \( C(3, 2) \):
\[ M = \left( \frac{1 + 3}{2}, \frac{8 + 2}{2} \right) = \left( 2, 5 \right) \]
The radius of the circumcircle is the distance from M to P (or M to A/B, since they are equidistant).
\[ \text{Radius} = \sqrt{(2 - 1)^2 + (5 - 8)^2} = \sqrt{1 + 9} = \sqrt{10} \]

6. **Write the Equation of the Circumcircle**:
The equation of the circle with center \( (h, k) = (2, 5) \) and radius \( r = \sqrt{10} \) is:
\[ (x - 2)^2 + (y - 5)^2 = 10 \]
Expanding this gives:
\[ x^2 - 4x + 4 + y^2 - 10y + 25 = 10 \]
\[ x^2 + y^2 - 4x - 10y + 19 = 0 \]

### Final Answer:
The equation of the circumcircle of triangle PAB is:
\[ x^2 + y^2 - 4x - 10y + 19 = 0 \]