The angle between the pair of tangents drawn from (1, 1) to the circle  $x^2+y^2+4x+4y-1=0$ is

Equation to the pair of tangents drawn from(1, 1) to the circle $x^2+y^2+4x+4y-1=0$ is

$\begin{array}{l}[x+y+2(x+1)+2(y+1)-1{]}^2=\left(1^2+1^2+4+4-1\right)\left(x^2+y^2+4x+4y-1\right)\\ \Rightarrow (3x+3y+3{)}^2=9\left(x^2+y^2+4x+4y-1\right)=(x+y+1{)}^2=x^2+y^2+4x+4y-1\\ \Rightarrow x^2+y^2+1+2x+2y+2xy=x^2+y^2+4x+4y-1\Rightarrow 2xy-2x-2y+2=0\end{array}$

$\Rightarrow xy-x-y+1=0.\therefore$Angle between the tangents = $\frac{\mathrm{\pi }}{2}$