Problem 14 - Algebra and Trigonometry

Question

\(3-16=\) Graph the inequality. $$ x^{2}+y^{2} \geq 9 $$

Step-by-Step Solution

Answer

The inequality represents a filled circle with a radius of 3 centered at the origin.

Step 1: Recognize the Equation Type

The given inequality is \( x^2 + y^2 \geq 9 \), which represents a geometrical shape in the Cartesian plane. Specifically, this is a circle inequality centered at the origin \((0, 0)\) with a radius of \(3\), derived from the standard form \( x^2 + y^2 = r^2 \).

Step 2: Identify the Region to Shade

Since the inequality is \( \geq \), you must include all points inside the circle and on the circle itself. This means we consider points where \( x^2 + y^2 \) is equal to or greater than \(9\).

Step 3: Sketch the Circle

Draw the circle with radius \(3\) centered at the origin. This is done by plotting the points at \((3,0)\), \((-3,0)\), \((0,3)\), and \((0,-3)\), and connecting them in a smooth round shape, thereby outlining the circle.

Step 4: Shade the Appropriate Region

Since we're dealing with \( \geq \), shade the entire area inside the circle all the way to the edge of the circle. This shading includes the boundary line (since it can be \(=9\) too).

Key Concepts

Circle Inequality

When we talk about a circle inequality such as \( x^2 + y^2 \geq 9 \), we are dealing with a particular type of geometric relation in a coordinate system. A circle equation is given by \( x^2 + y^2 = r^2 \), where \( r \) is the radius of the circle. An inequality, such as \( \geq \), indicates the area we are interested in, which includes the circle itself and the surrounding region outwards. This is different from \( < \), which would consider only the inside without the border. Here, our circle is centered at the origin \((0, 0)\) with a radius \( r = 3 \). This results from the expression \( r^2 = 9 \). By understanding the inequality sign we clarify the scope of points included in the solution set.

Cartesian Plane

The Cartesian plane is a two-dimensional plane defined by two perpendicular axes: the x-axis and the y-axis. Each point on this plane is identified by a pair of numerical coordinates \((x, y)\). The Cartesian plane is crucial for graphing inequalities because it gives a visual representation of mathematical equations and inequalities. In the case of our circle inequality, \( x^2 + y^2 \geq 9 \), the Cartesian plane allows us to plot the circle centered at \((0, 0)\). On this plane, each point that satisfies the inequality will either lie on the circle or outside it, helping us visualize the solution.

Geometric Shapes

In this context, the primary geometric shape is a circle, defined by its standard equation \( x^2 + y^2 = r^2 \). A circle is a set of all points in a plane that are a constant distance from a given point, the center. Here, the center is \((0, 0)\), and each point on the circle is exactly \(3\) units away from the origin. Circles are an essential shape in geometry, and understanding their representation and properties is key to graphing them correctly in mathematical problems. For inequalities, we often modify how these shapes visually represent data, showing areas inside, on, or outside the shapes, depending on the inequality.

Inequality Shading

Inequality shading is a technique used in graphing to indicate the region that satisfies an inequality. When we graph \( x^2 + y^2 \geq 9 \), we are not only interested in the points exactly on the circle but also in all the points surrounding and including it. This means drawing a circle to represent where \( x^2 + y^2 = 9 \) would be, and then shading the area inside and extending outward from that circle. This shaded area shows all solutions to the inequality. Inequality shading visually differentiates the solution set from non-solutions, allowing us to understand the scope of possible values that satisfy the inequality. Including the boundary in \( \geq \) or \( \leq \) is an important detail, as it confirms whether the points on the line are solutions or not. In this exercise, the shaded area inside the circle up to its boundary confirms that all these points satisfy the given inequality.