IIT JEE Advanced 2009 Paper 1

Section: Mathematics

Question : 48 of 60

Marks: +1, -0

Tangents drawn from the point P(1, 8) to the circle x2+y2 − 6x − 4y − 11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is

[JEE Adv 2009 P1]

x2+y2 + 4x - 6 y +19 = 0
x2+y2 - 4x -10 y +19 = 0
x2+y2 - 2x + 6 y - 29 = 0
x2+y2 - 6x - 4 y +19 = 0

Validate

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