Problem 13 - College Algebra

Question

Graph each inequality. $$ x^{2}+y^{2} \leq 1 $$

Step-by-Step Solution

Answer

Graph a circle with center at the origin and radius of 1 and then shade the area inside and the circle itself

Step 1: Identifying the inequality

Firstly, identify the inequality given. In this case, the inequality is \(x^{2}+y^{2} \leq 1\). This represents the equation of a circle with radius 1 and the center at the origin (0,0). The inequality sign '\(\leq\)' tells us that the solution includes all points inside the circle and on the circle itself.

Step 2: Plotting the circle

Now, you should sketch the circle using the given information. The origin (0,0) serves as the center of the circle. As the radius is 1 unit, points on the circle will be distanced exactly 1 unit away from the center in all directions.

Step 3: Shading the inequality area

Since the inequality sign is '\(\leq\)', all the points inside and on the circle will be a solution to the given inequality. Therefore, shade the area inside and the circle itself, this represents the points that satisfy the inequality.

Key Concepts

Circle Equations

A circle equation in standard form is typically written as \(x^2 + y^2 = r^2\), where \(r\) represents the radius of the circle and \((0, 0)\) is the center.
In this particular problem, we have \(x^2 + y^2 \leq 1\), which still takes the form of a circle equation, with a twist. The inequality sign "\(\leq\)" indicates that the points we are interested in are not just on the circle itself but also inside it.
In interpreting circle equations in algebraic contexts, inequalities enrich our understanding by broadening the geometry to regions inside or outside circles.
This knowledge is pivotal as it allows us to visualize not just boundary lines but also broader areas defined by the equation.

Radius and Center

The radius and center are fundamental aspects of any circle equation. In the equation \(x^2 + y^2 = r^2\), the center

• Is usually given as \((h, k)\). In simplified cases like \(x^2 + y^2 = 1\), the center is \((0, 0)\), meaning the circle is centered at the origin.

The radius in this equation is determined by taking the square root of the right-hand side, which is 1 here, meaning:

• The radius is 1 unit.

Understanding these parameters equips you to quickly visualize the circle on a graph, knowing exactly where it’s centered and how far it stretches from that center point.
These two elements, radius and center, will anchor your graph, aiding in plotting it accurately.

Plotting Points

Plotting points for the equation of a circle starts by identifying points at a distance equivalent to the radius from the center.

• To plot \(x^2 + y^2 = 1\), start at the center (0,0).
• Mark points 1 unit away horizontally \((1, 0)\) and \((-1, 0)\).
• Extend the same logic vertically with \((0, 1)\) and \((0, -1)\).

These initial points help outline the circle’s symmetric nature around the center.
Additionally, by connecting these points smoothly, you form a circle. For the inequality \(x^2 + y^2 \leq 1\), consider the entire shaded region within and including this circular boundary, ensuring no points are missed.

Shading Regions

Shading regions in graphs involving inequalities is crucial to distinguishing solutions inclusive of areas within specified boundaries.
When handling \(x^2 + y^2 \leq 1\):

• The inequality indicates that not only the boundary (circle) but also the interior is part of the solution set.
• Use shading or coloring inside the circle, representing every point within the circle as a valid solution for the inequality.

By shading these regions, one demonstrates visually how points satisfy the inequality, reinforcing the notion that all shaded areas meet the criteria of the inequality rather than merely the boundary line.
This step ties together the theory with tangible visuals, making it easier for students to see the area they are working with.