1. We are given two inequalities: $$x^2 + y^2 \leq 4$$ and $$ (x + 2)^2 + y^2 \leq 4.$$ Our goal is to understand the region defined by these inequalities. 2. The inequality $$x^2 + y^2 \leq 4$$ represents a circle centered at the origin $$(0,0)$$ with radius 2. 3. The inequality $$ (x + 2)^2 + y^2 \leq 4$$ represents a circle centered at $$(-2, 0)$$ with radius 2. 4. To find the region common to both inequalities, we look at the overlap of these two circles. 5. Both circles have radius 2, with centers 2 units apart horizontally. 6. The region $$x^2 + y^2 \leq 4$$ includes all points within 2 units of the origin. 7. The region $$ (x + 2)^2 + y^2 \leq 4$$ includes all points within 2 units of $(-2,0)$. 8. The solution set consists of all points inside or on both circles, meaning the intersection of these two disks. Final answer: The system $$x^2 + y^2 \leq 4, \quad (x + 2)^2 + y^2 \leq 4$$ defines the overlapping region of two circles each of radius 2, centered at $(0,0)$ and $(-2,0)$ respectively.