Tangents drawn from the point $P (1, 8)$ to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the points $A$ and $B .$ The equation of the circumcircle of the triangle in $P A B$ is

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$x^{2} + y^{2} - 6 x - 4 y + 19 = 0$

$x^{2} + y^{2} - 2 x + 6 y - 29 = 0$

$x^{2} + y^{2} + 4 x - 6 y + 19 = 0$

$x^{2} + y^{2} - 4 x - 10 y + 19 = 0$

answer is B.

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Detailed Solution

Equation of $A B$ the Chord of contact of $p$ is

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

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

Equation of any circle through. $A B$ is

$x^{2} + y^{2} - 6 x - 4 y - 11 + λ (x - 3 y + 15) = 0$

It will pass through $P (1, 8)$ if

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

Thus, equation of the required circle is

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

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