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Tangents are drawn from the point $C (1, 8)$ to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0 .$ If $A B$ is the chord of contact, then the circumcircle of triangle $A B C$ 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

Centre $= O (3, 2)$
$∠ O B C = ∠ O A C = \frac{π}{2}$

The circle described on $O C$ as diameter passes through $A$ and $B$ .
$∴$ The desired circle is
$(x - 3) (x - 1) + (y - 2) (y - 8) = 0$
or $x^{2} + y^{2} - 4 x - 10 y + 19 = 0$ .

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