The length of the tangent drawn to the circle $x^2+y^2-2 x+4 y-11=0$ from the point $(1,3)$ is :
The length of the tangent drawn to the circle $x^2+y^2-2 x+4 y-11=0$ from the point $(1,3)$ is :
1
2
3
4
Solution
The length of the tangent drawn to the circle $x^2+y^2-2 x+4 y-11=0$ from the point $(1,3)$
$=\sqrt{1^2+3^2-2 \cdot 1+12-11}$
$=\sqrt{1+9-2+12-11}$
$=\sqrt{22-13}=\sqrt{9}$
$=3$