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

Question

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 (A) x2+y2+4x-6y+19=0 (B) x2+y2-4x-10y+19=0 (C) x2+y2-2x+6y-29=0 (D) x2+y2-6x-4y+19=0

Tardigrade Admin · Accepted Answer

Using the property that given point P(.1,8.) and centre of the given circle (.3,2.) are ends of diameter of the required circumcircle of Δ PAB . So using diametric form, we get required equation (x - 1)(x - 3)+(y - 8)(y - 2)=0 i.e. x2+y2-4x-10y+19=0

Q. 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 the points $A$ and $B$ . The equation of the circumcircle of the triangle $P A B$ is

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

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

$x^{2} + y^{2} - 2 x + 6 y - 29 = 0$

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

Using the property that given point $P (1, 8)$ and centre of the given circle $(3, 2)$ are ends of diameter of the required circumcircle of $Δ P A B$ .
So using diametric form, we get required equation $(x - 1) (x - 3) + (y - 8) (y - 2) = 0$
i.e. $x^{2} + y^{2} - 4 x - 10 y + 19 = 0$

Q. 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

x2+y2+4x−6y+19=0

x2+y2−4x−10y+19=0

x2+y2−2x+6y−29=0

x2+y2−6x−4y+19=0