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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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

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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

The correct option is B x2+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)=0or x2+y2−4x−10y+19=0

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

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
$(x - 3) (x - 1) + (y - 2) (y - 8) = 0 o r x^{2} + y^{2} - 4 x - 10 y + 19 = 0$