$S \equiv x^2+y^2-2 x-4 y-4=0$ and $S^{\prime} \equiv x^2+y^2-4 x-2 y-16=0$ are two circles the point $(-2,-1)$ lies

Let the given circles be $S(x, y) = x^2 + y^2 - 2x - 4y - 4 = 0$ and $S'(x, y) = x^2 + y^2 - 4x - 2y - 16 = 0$.

To determine the position of the point $P(-2, -1)$ with respect to circle $S$, substitute the coordinates of $P$ into the expression for $S$:

$S(-2, -1) = (-2)^2 + (-1)^2 - 2(-2) - 4(-1) - 4 = 4 + 1 + 4 + 4 - 4 = 9$.

Since $S(-2, -1) = 9 > 0$, the point $P(-2, -1)$ lies outside the circle $S$.

To determine the position of the point $P(-2, -1)$ with respect to circle $S'$, substitute the coordinates of $P$ into the expression for $S'$:

$S'(-2, -1) = (-2)^2 + (-1)^2 - 4(-2) - 2(-1) - 16 = 4 + 1 + 8 + 2 - 16 = -1$.

Since $S'(-2, -1) = -1 < 0$, the point $P(-2, -1)$ lies inside the circle $S'$.

Therefore, the point lies inside $S'$ only.

Answer: inside $S^{\prime}$ only