The equation of the pair of tangents drawn from the point $(1,1)$ to the circle $x^2+y^2+2 x+2 y+1=0$ is
The equation of the pair of tangents drawn from the point $(1,1)$ to the circle $x^2+y^2+2 x+2 y+1=0$ is
$3 x^2-8 x y+3 y^2-2 x-2 y+6=0$
$11 x^2-8 x y+11 y^2-4 x-4 y-6=0$
$3 x^2-8 x y+3 y^2+2 x+2 y-2=0$
$x^2-4 x y+y^2+x+y=0$
Solution
Equation of pair of tangents from point $(1,1)$ on circle $S=x^2+y^2+2 x+2 y+1$ can be given as,
$
\begin{aligned}
& S S_1=T^2 \\
& \Rightarrow\left(x^2+y^2+2 x+2 y+1\right)\left(1^2+1^2+2+2+1\right)
\end{aligned}
$
$\begin{aligned} & =[x(1)+y(1)+(x+1)+(y+1)+1]^2 \\ & \Rightarrow 7 x^2+7 y^2+14 x+14 y+7=(2 x+2 y+3)^2 \\ & \Rightarrow 3 x^2+3 y^2+2 x+2 y-2-8 x y=0\end{aligned}$