A geometry problem by Archit Tripathi

Using the concept of family of curves $H + u(L)^{2} = 0$

This will give the equation of all the curves which pass through the points of intersection of these curves.

$\Rightarrow$ $(x^{2} + y^{2} + 4xy + 8x + 8y + 8) + u(x + y + 1)^{2} = 0$

$\Rightarrow$ $(1 + u)(x^{2} + y^{2}) +(4 + 2u)xy + (8 + 2u)(x + y) + (8 + u) = 0$

Now if we want this curve to be a circle

Coefficient of $XY = 0$

$\Rightarrow$ $4 + 2u = 0$

$\Rightarrow$ $u = -2$

The equation of the circle comes out to be

$x^{2} + y^{2} - 4x - 4y - 6 = 0$

whose radius is $R^{2} = 4 + 4 - (-6) = \boxed{14}$