Tangents are 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 point A and B, then equation of the circumcircle of the triangle PAB is

x^{2} + y^{2} + 4 x - 6 y + 9 = 0

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

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

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

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Solution

The correct option is B $x^{2} + y^{2} - 4 x - 10 y + 19 = 0$
Given equation of circle is $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$
Centre (C) = (3,2), radius = $\sqrt{24}$
Centre of circum-circle=mid point of PAC = (2,5)
Radius $= \frac{1}{2} (P C) = \frac{1}{2} \sqrt{4 + 36} = \sqrt{10}$
$∴$ equation of circum circle is $(x - 2)^{2} + (y - 5)^{2} = 10 i . e . x^{2} + y^{2} - 4 x - 10 y + 19 = 0$

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