Q. 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
A) $x^{2} + y^{2} + 4x - 6y + 19 = 0$
B) $x^{2} + y^{2} - 4x - 10y + 19 = 0$
C) $x^{2} + y^{2} - 2x + 6y - 29 = 0$
D) $x^{2} + y^{2} - 6x - 4y + 19 = 0$
Explanation
(b)
The centre of the given circle is $O(3,\ 2)$
[image:11_image10.emf]
Since, $OA$ and $OB$ are perpendicular to $PA$and $PB$.
Also, $OP$ is the diameter of the circumcircle of $\bigtriangleup PAB$
Its equation is
$(x - 3)(x - 1) + (y - 2)(y - 8) = 0$
$\Rightarrow \ \ \ \ x^{2} + y^{2} - 4x - 10y + 19 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.