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 in $PAB$ is

Equation of $AB$ the Chord of contact of $p$ is

$\begin{array}{l}1\cdot x+8y-3(x+1)-2(y+8)-11=0\\ \Rightarrow -2x+6y-30=0\end{array}$

$\Rightarrow x-3y+15=0$

Equation of any circle through. $AB$ is

          $x^2+y^2-6x-4y-11+\lambda (x-3y+15)=0$

It will pass through $P(1,8)$ if

$\begin{array}{l} 1+64-6-32-11+\lambda (1-24+15)=0\\ \Rightarrow 16-8\lambda =0\Rightarrow \lambda =2\end{array}$

Thus, equation of the required circle is

$\begin{array}{l} x^2+y^2-6x-4y-11+2(x-3y+15)=0\\ \Rightarrow x^2+y^2-4x-10y+19=0\end{array}$