(x-2)2 + (y+2)2 - 9 >= 0

To analyze the inequality \((x-2)^2 + (y+2)^2 - 9 \geq 0\), we can first rewrite it in a more familiar form. 1. Start by rearranging the inequality: \[ (x-2)^2 + (y+2)^2 \geq 9 \] 2. This expression represents the equation of a circle centered at the point \((2, -2)\) with a radius of \(3\) (since \(9\) is \(3^2\)). The inequality states that the points \((x, y)\) must lie outside the circle or on the circle itself. 3. To further clarify, you can express it as: \[ \sqrt{(x-2)^2 + (y+2)^2} \geq 3 \] This tells us that the distance from the point \((x, y)\) to the center \((2, -2)\) is at least \(3\). ### Summary The solution to the inequality \((x-2)^2 + (y+2)^2 - 9 \geq 0\) consists of all points \((x, y)\) that satisfy the condition of being outside or on the circle of radius \(3\) centered at \((2, -2)\). These points can be described in set notation as: \[ \{(x, y) \mid \sqrt{(x-2)^2 + (y+2)^2} \geq 3\} \]