Q26 / 60 · JEE Advanced 2009 (Paper 1) · Mathematics

Tangents drawn from the point $P(1,8)$ to the circle $x^2+y^2-6 x-4 y-11=0$ touch the circle at the points $A$ and $B$. The equation of the circumcircle of $\triangle P A B$ is

  • A
    $x^2+y^2+4 x-6 y+19=0$
  • Correct answer
    B
    $x^2+y^2-4 x-10 y+19=0$
  • C
    $x^2+y^2-2 x+6 y-29=0$
  • D
    $x^2+y^2-6 x-4 y+19=0$

Solution

For required circle, $P(1,8)$ and $O(3,2)$ will be the end point of its diameter. $$ \begin{aligned} & \therefore(x-1)(x-3)+(y-8)(y-2)=0 \\ & \Rightarrow \quad x^2+y^2-4 x-10 y+19=0 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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