question_answer

Tangents drawn from the point P(1, 8) to the circle \[{{\operatorname{x}}^{2}}+{{y}^{2}}- 6x -4y -11 = 0\] 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) \[{{x}^{2}}+{{y}^{2}}-6x-4y+19=0\]

Correct Answer: B

Solution :

Let, centre of given circle \[{{x}^{2}}+{{y}^{2}}-6x-4y-11=0\] is 0(3, 2). For required circle, P(1, 8) and 0(3, 2) will be the end points of its diameter. \[\therefore \,\,\,\left( x-1 \right)\left( x-3 \right)+\left( y-8 \right)\left( y-2 \right)=0\]   [From figure \[{{\operatorname{OP}}^{2}}-P{{A}^{2}}= Radius of\,\,circumcircle\] so P and O are end point of diameter] \[\Rightarrow \,\,\,{{x}^{2}}+{{y}^{2}}-4x-10y+19=0\]c