question_answer

Tangent drawn from the point (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 \[\Delta \,PAB\] 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 :

Equation of circle

\[{{x}^{2}}+{{y}^{2}}-6x-4y-11=0\]

Since, OA and OB is perpendicular to PA and PB

OP diameter of the circumcircle of\[\Delta PAB\]equation of circle is

\[(x-3)(x-1)+(y-2)(y-8)=0\]

\[\Rightarrow \]\[{{x}^{2}}-4x+3+{{y}^{2}}-10y+16=0\]

\[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-4x-10y+19=0\]