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 the triangle $P A B$ is

  • [IIT 2009]
  • A

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

  • B

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

  • C

    $x^2+y^2-2 x+6 y-29=0$

  • D

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

Similar Questions

The equation of the tangent to the circle ${x^2} + {y^2} = {r^2}$ at $(a,b)$ is $ax + by - \lambda = 0$, where $\lambda $ is

The equation of the tangent to the circle ${x^2} + {y^2} = {a^2}$ which makes a triangle of area ${a^2}$ with the co-ordinate axes, is

Let the tangent to the circle $C _{1}: x^{2}+y^{2}=2$ at the point $M (-1,1)$ intersect the circle $C _{2}$ : $( x -3)^{2}+(y-2)^{2}=5$, at two distinct points $A$ and $B$. If the tangents to $C _{2}$ at the points $A$ and $B$ intersect at $N$, then the area of the triangle $ANB$ is equal to

  • [JEE MAIN 2022]

Let $B$ be the centre of the circle $x^{2}+y^{2}-2 x+4 y+1=0$ Let the tangents at two points $\mathrm{P}$ and $\mathrm{Q}$ on the circle intersect at the point $\mathrm{A}(3,1)$. Then $8.$ $\left(\frac{\text { area } \triangle \mathrm{APQ}}{\text { area } \triangle \mathrm{BPQ}}\right)$ is equal to .... .

  • [JEE MAIN 2021]

If the centre of a circle is $(-6, 8)$ and it passes through the origin, then equation to its tangent at the origin, is