Problem 17 - College Algebra

Question

In Exercises 1–26, graph each inequality. $$(x-2)^{2}+(y+1)^{2}<9$$

Step-by-Step Solution

Answer

The graph of the inequality \((x-2)^{2}+(y+1)^{2}<9\) shows a shaded region inside a circle centered at (2,-1) with a radius of 3 units. The shaded region, excluding the border of the circle, represents the solution set of the inequality.

Step 1: Identify the center and radius

By comparing the inequality \((x-2)^{2}+(y+1)^{2}<9\) with the standard equation of a circle \((x-h)^2 + (y-k)^2 = r^2\), we can identify (h, k) as the center of the circle and r to be the radius, where h = 2, k = -1 and \( r = \sqrt{9} = 3 \). So, we have the center at (2,-1) and radius as 3.

Step 2: Draw the circle

Draw a circle with the identified the center and radius. This is done by placing a point at the center (2, -1), then using this point as a center and your radius length (3 units), draw the circle.

Step 3: Identify solution inside the circle

Because the inequality is less than and not less than or equal to, the solution set consists of all points inside the circle, not including the points on the circle.

Step 4: Shade the interior of the circle

Shade the interior of the circle. This shaded region is the solution area for this inequality. Any point in this area will satisfy the inequality.

Key Concepts

Circle Equations

Understanding the basic form of a circle equation is essential when tackling problems involving graphing inequalities. A circle's equation in the standard form is given by \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h, k) \) is the center of the circle, and \( r \) is the radius. When graphing an inequality like \( (x-2)^2 + (y+1)^2 < 9 \) we're interested in the region it represents.
Crucially, the inequality sign dictates that we are looking for points within the circle, excluding the boundary. This inequality essentially forms a 'boundary-less' circle, where every coordinate that satisfies the inequality is located strictly inside the actual circle's edge. This concept is pivotal because it illustrates the set of points that create the solution.
Graphing such inequalities requires careful plotting of the center of the circle, followed by drawing a curve representing the locus points equidistant from the center, with the understanding that the actual boundary is not part of the solution.

Inequality Solutions

The solutions to an inequality are the set of all points that make the inequality true. In the case of the inequality \( (x-2)^{2}+(y+1)^{2}<9 \) our solution set is not a line or a single point, but an area. This is because the inequality involves two variables leading to a two-dimensional solution region.
For inequalities like this, where we have a '<' sign instead of a '\leq', the region of the graph that represents the solution is open, meaning it does not include the line—or in our case, the circle—itself. The reason for this is tied to the strict '<' inequality: points on the circle would result in the expression equating to 9, not less than it. Thus, we shade the interior of the circle exclusively to visually represent this solution area, leaving the circle boundary unmarked.

Radius and Center of a Circle

The radius and the center are two attributes that define a unique circle on a Cartesian plane. The radius is the constant distance from the center of the circle to any point on the circle's edge. When we are given a standard form equation of a circle, \( (x-h)^2 + (y-k)^2 = r^2 \), we can extract the center's coordinates as \( (h, k) \) and the radius as the square root of \( r^2 \), which is \( r \).
Applying this knowledge to our inequality exercise, we determined that the center is at \( (2, -1) \) and the radius is 3 units. When graphing, these values allow us to accurately place the center of the circle and use a compass or a ruler to draw a circle at a consistent distance from this center point. It's the precise combination of accurate center placement and radius measurement that ensures our graphed circle represents the true solution set for the given inequality.