Coordinate Geometry

Let us write the required midpoint formulae as:

$x = \frac{-3 + k}{2}; y = \frac{2 + [1 \pm \sqrt{4 - (k+3)^2}]}{2}$

where $k \in \mathbb{R}$ . If we solve for $k$ in terms of $x$ in the first equation and then substitute it into the second, we now end up with:

$y = \frac{2 + [1 \pm \sqrt{4 - ((2x+3) + 3)^2}]}{2} = \frac{3 \pm \sqrt{4 - (2x+6)^2}}{2}$ ;

or $2y - 3 = \sqrt{4 - 4(x+3)^2}$ ;

or $(2y-3)^2 = 4 - 4(x+3)^2;$

or $4(y- \frac{3}{2})^2 = 4 - 4(x+3)^2;$

or $\boxed{(x+3)^2 + (y - \frac{3}{2})^2 = 1}$

So the locus of $G$ is just a unit circle $\Rightarrow$ the maximum distance between any two points $\in G = \boxed{2}.$