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AIEEE 2009

JEE Main 2009 Mathematics question on Circles

Tangents drawn from the point P(1,8)P(1, 8) to the circle x2+y2−6x−4y−11=0x^2 + y^2 - 6x - 4y - 11 = 0 touch the circle at the points AA and BB. The equation of the circumcircle of △PAB\triangle PAB is
  1. A.
    x2+y2+4x−6y+19=0x^2 + y^2 + 4x - 6y + 19 = 0
  2. B.
    x2+y2−4x−10y+19=0x^2 + y^2 - 4x - 10y + 19 = 0
  3. C.
    x2+y2−2x+6y−29=0x^2 + y^2 - 2x + 6y - 29 = 0
  4. D.
    x2+y2−6x−4y+19=0x^2 + y^2 - 6x - 4y + 19 = 0
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Correct option: B

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