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

Given equation of circle is $x^2+y^2+4x+4y-1=0$

Centre$\left(-2,-2\right)$

$OA=OB=r=\sqrt{4+4+1}=3$ Where $r$ is radius of circle.

$OC=\sqrt{9+9}=3\sqrt{2}$ (by distance formula)

$\Rightarrow O{C}^2=O{A}^2+C{A}^2$(Pythagoras theorem)

$\Rightarrow C{A}^2=9$

Now, $\sin \alpha =\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos \alpha =\frac{1}{\sqrt{2}}$

$\Rightarrow \sin 2\alpha =2\sin \alpha \cos \alpha =1$

$\Rightarrow \sin 2\alpha =\sin \frac{\mathrm{\pi }}{2}$

$\Rightarrow 2\alpha =\frac{\mathrm{\pi }}{2}$

So $\frac{\mathrm{\pi }}{2}$ is the required angle between the pair of tangents.