Question

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

Answer 1

The correct option is B $x^2+y^2-4x-10y+19=0$

Note: Circumcircle of triangle $PAB$ will always pass through the centre of the circle on which $A$ and $B$ are lying i.e. the circle to which tangents are drawn from $P$.

Note: Also $PO$ acts as the diameter of the circumcircle.

$\therefore$ Equation of the circumcircle:

$\Rightarrow (x-1)(x-3)+(y-8)(y-2)=0$

$\Rightarrow x^2-3x-x+3+y^2-2y-8y+16=0$

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

$\therefore B)$ Answer.