Find Circumcircle of Triangle Formed by Tangents to Circle

To find the equation of the circumcircle of $\triangle PAB$, let us consider the geometric properties of tangents drawn to a circle:

1. Center of the given circle:
   The equation of the circle is given as:
   $x^2 + y^2 - 6x - 4y - 11 = 0$
   Comparing this with the standard equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, we find:
   $2g = -6 \implies g = -3$
   $2f = -4 \implies f = -2$
   Thus, the center of the circle is $C(3, 2)$.

2. Geometric Properties:
   Let $C$ be the center of the given circle, and let the tangents from the external point $P(1, 8)$ touch the circle at points $A$ and $B$.
   Since the tangent at any point on a circle is perpendicular to the radius through the point of contact, we have:
   $\angle PAC = 90^\circ \quad \text{and} \quad \angle PBC = 90^\circ$

   In the quadrilateral $PACB$, the sum of the opposite angles is:
   $\angle PAC + \angle PBC = 90^\circ + 90^\circ = 180^\circ$
   Therefore, the quadrilateral $PACB$ is concyclic, meaning that the four points $P, A, C,$ and $B$ lie on a single circle.

3. Circumcircle of $\triangle PAB$:
   Since $\angle PAC = 90^\circ$ and $\angle PBC = 90^\circ$, the line segment $PC$ subtends a right angle at both $A$ and $B$. By Thales's theorem, $PC$ must be a diameter of the circle passing through $P, A, C,$ and $B$.
   Consequently, the circumcircle of $\triangle PAB$ is the circle having the line segment $PC$ as its diameter.

4. Equation of the Circle:
   The diametric form of the equation of a circle passing through $P(x_1, y_1) = (1, 8)$ and $C(x_2, y_2) = (3, 2)$ is:
   $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$

   Substituting the coordinates:
   $(x - 1)(x - 3) + (y - 8)(y - 2) = 0$
   $(x^2 - 4x + 3) + (y^2 - 10y + 16) = 0$
   $x^2 + y^2 - 4x - 10y + 19 = 0$

Therefore, the correct option is B.