In Exercises 27–62, graph the solution set of each system of inequalities or indicate that the system has no solution. x²+y²>1, x²+y²<16

Verified step by step guidance

1. Step 1: Understand the inequalities. The first inequality is $x^2+y^2>\; 9$, which represents all points outside the circle centered at the origin with radius 3 (since 9 is 3 squared).
2. Step 2: The second inequality is , which represents all points inside the circle centered at the origin with radius 5 (since 25 is 5 squared).
3. Step 3: To find the solution set of the system, identify the region where both inequalities are true simultaneously. This means the points must lie outside the smaller circle (radius 3) and inside the larger circle (radius 5).
4. Step 4: Graphically, this solution set is the annular region (ring-shaped area) between the two circles, excluding the boundaries because the inequalities are strict (greater than and less than, not greater than or equal to or less than or equal to).
5. Step 5: When graphing, draw the circle with radius 3 as a dashed circle to indicate points on the circle are not included, and similarly draw the circle with radius 5 as a dashed circle. Shade the region between these two circles to represent the solution set.

Key Concepts

Inequalities Involving Circles

Inequalities like x² + y² > r² and x² + y² < R² represent regions outside or inside circles centered at the origin with radii r and R, respectively. Understanding these inequalities helps identify which points satisfy the conditions relative to the circles.

Graphing Solution Sets of Systems of Inequalities

Graphing systems of inequalities involves shading regions that satisfy all inequalities simultaneously. For two inequalities involving circles, the solution set is the area where the shaded regions overlap, which can be inside one circle but outside another.

Set Intersection and No Solution Scenarios

The solution to a system of inequalities is the intersection of their individual solution sets. If the regions do not overlap, the system has no solution. Recognizing when inequalities represent disjoint regions is key to determining if solutions exist.