Tangents drawn from the point P(1,8) to the c | vedclass

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(B) The given circle is $x^2+y^2-6x-4y-11=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get the center $C(-g, -f) = (3, 2)$.Si

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

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(B) The given circle is $x^2+y^2-6x-4y-11=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get the center $C(-g, -f) = (3, 2)$.Since $PA$ and $PB$ are tangents from point $P(1,8)$ to the circle,the radii $CA$ and $CB$ are perpendicular to the tangents at points $A$ and $B$ respectively. Thus,$\angle PAC = 90^\circ$ and $\angle PBC = 90^\circ$.This implies that the points $A$ and $B$ lie on a circle with $PC$ as the diameter. The triangle $PAB$ is inscribed in this circle,so the circumcircle of $	riangle PAB$ is the circle with diameter $PC$.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Using $P(1,8)$ and $C(3,2)$ as diameter endpoints:$(x-1)(x-3) + (y-8)(y-2) = 0$$x^2 - 4x + 3 + y^2 - 10y + 16 = 0$$x^2 + y^2 - 4x - 10y + 19 = 0$.Thus,the correct option is $B$.

Tangents drawn from the point $P(1,8)$ to the circle $x^2+y^2-6x-4y-11=0$ touch the circle at the points $A$ and $B$. The equation of the circumcircle of the triangle $PAB$ is

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