Tangents drawn from the point P(1, 8) to the circle x2+y2−6x−4y−11=0 touch the circle at points A and B. The equation of the circumcircle of triangle PAB is
A
x2+y2+4x−6y+19=0
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B
x2+y2−4x−10y+19=0
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C
x2+y2−2x+6y−20=0
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D
x2+y2−6x−4y+19=0
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Solution
The correct option is Bx2+y2−4x−10y+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)=0orx2+y2−4x−10y+19=0