Question

Tangents drawn from the point $P\left(1 , 8\right)$ 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

Answer: (B)

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