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 points A and B. The equation of the circumcircle of triangle PAB is

x^{2} + y^{2} + 4 x - 6 y + 19 = 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 - 20 = 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 $x^{2} + y^{2} - 4 x - 10 y + 19 = 0$

The centre of the circle is C(3, 2).
Since CA and CB are prependicular to PA and PB, CP is the diameter of the circumcircle of triangle PAB. Its equation is
$(x - 3) (x - 1) + (y - 2) (y - 8) = 0 o r x^{2} + y^{2} - 4 x - 10 y + 19 = 0$

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x^{2} + y^{2} - 6 x - 4 y - 11 = 0

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