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

$\text{Circle: } x^{2} + y^{2} - 6x - 4y - 11 = 0$ $\text{Point: } P(1, 8)$ $\text{To find: equation of the circumcircle of } \triangle PAB$ $1.\ \text{centre } C \text{ and radius } r \text{ of the given circle}$ $2.\ \text{tangent } \perp \text{radius} \Rightarrow \angle CAP = \angle CBP = 90^{\circ}$ $3.\ \text{circumcircle of } \triangle PAB \text{, then match the option}$ $x^{2} + y^{2} - 6x - 4y - 11 = 0$ $\Rightarrow (x-3)^{2} - 9 + (y-2)^{2} - 4 - 11 = 0$ $\Rightarrow (x-3)^{2} + (y-2)^{2} = 24$ $C(3, 2), \quad r = \sqrt{24} = 2\sqrt{6}$ $PC = \sqrt{(3-1)^{2} + (2-8)^{2}} = \sqrt{40} = 2\sqrt{10}$ $PC = 2\sqrt{10} > r = 2\sqrt{6} \Rightarrow P \text{ is outside the circle}$ $\text{Tangent} \perp \text{radius} \Rightarrow \angle CAP = \angle CBP = 90^{\circ}$ $\text{By Thales: } \angle CAP = 90^{\circ} \Rightarrow A \text{ on circle with diameter } PC$ $\text{Circumcircle of } \triangle PAB = \text{circle with diameter } PC$ $C \text{ is also on the circumcircle (other end of diameter } PC\text{)}$ $M = \left(\dfrac{1+3}{2},\ \dfrac{8+2}{2}\right) = (2, 5)$ $R = \dfrac{PC}{2} = \dfrac{2\sqrt{10}}{2} = \sqrt{10}$ $(x-2)^{2} + (y-5)^{2} = 10$ $x^{2} + y^{2} - 4x - 10y + (4 + 25 - 10) = 0$ $x^{2} + y^{2} - 4x - 10y + 19 = 0$ $P(1, 8): \ 1 + 64 - 4 - 80 + 19 = 0 \ \checkmark$ $\boxed{\text{Answer: (2)}\ x^{2} + y^{2} - 4x - 10y + 19 = 0}$