Problem 63 - Intermediate Algebra

Question

Write each equation in standard form, if it is not already so, and graph it. If the graph is a circle, give the coordinates of its center and its radius. If the graph is a parabola, give the coordinates of its vertex. $$ x^{2}+y^{2}-6 x+8 y+18=0 $$

Step-by-Step Solution

Answer

The equation represents a circle with center (3, -4) and radius \(\sqrt{7}\).

Step 1: Identify the Given Equation Type

The given equation is \(x^2 + y^2 - 6x + 8y + 18 = 0\). The presence of both \(x^2\) and \(y^2\) terms with equal coefficients suggests that the equation might represent a circle.

Step 2: Rearrange and Complete the Square

First, rearrange the equation to group \(x\) and \(y\) terms: \(x^2 - 6x + y^2 + 8y = -18\). Now, complete the square for \(x\) and \(y\) separately: \(\text{For } x: x^2 - 6x = (x-3)^2 - 9 \\text{For } y: y^2 + 8y = (y+4)^2 - 16\). Substitute back: \((x-3)^2 - 9 + (y+4)^2 - 16 = -18\).

Step 3: Simplify the Equation

Simplify the equation from the previous step: \((x-3)^2 + (y+4)^2 - 25 = -18\). Add 25 to both sides to get the standard form: \((x-3)^2 + (y+4)^2 = 7\).

Step 4: Identify the Circle's Characteristics

From the equation \((x-3)^2 + (y+4)^2 = 7\), we see it's in the standard form \((x-h)^2 + (y-k)^2 = r^2\). Identify the center \((h, k)\) as \((3, -4)\) and the radius \(r = \sqrt{7}\).

Step 5: Graph the Circle

Plot the center of the circle at \((3, -4)\) in a coordinate plane. Draw the circle with radius \(\sqrt{7}\). The circle extends approximately 2.65 units from the center in all directions.

Key Concepts

Standard Form

The standard form of a circle's equation is crucial as it simplifies the process of identifying key circle attributes like the center and radius. The standard form is: \[(x-h)^2 + (y-k)^2 = r^2\] Here,

• \( (h, k) \) represents the center of the circle.
• \( r \) is the radius.

In the given problem, the initial equation was \( x^2 + y^2 - 6x + 8y + 18 = 0 \), which is not in the standard form at first glance. To transition this to the standard form, rearrangement and algebraic manipulation are necessary, as shown in the step-by-step solution. By completing the square and simplifying, we can express the equation in the form \( (x-3)^2 + (y+4)^2 = 7 \), which perfectly matches the standard form. Understanding this form allows us to quickly find the center and radius of the circle, which is essential for graph plotting and analysis.

Completing the Square

Completing the square is a powerful algebraic technique used to convert a quadratic expression into a form that can easily reveal geometric properties, like circles and parabolas. This technique involves the following steps:

• Rearranging the equation to group similar terms.
• Adding and subtracting the same value to create a perfect square trinomial.

In our exercise, we start with \( x^2 - 6x + y^2 + 8y = -18 \). To complete the square for the \( x \)-terms:

• Take half of \( -6 \), the coefficient of \( x \), to get \( -3 \), then square it to find \( 9 \).
• Rewrite \( x^2 - 6x \) as \((x-3)^2 - 9 \).

For the \( y \)-terms:

• Take half of \( 8 \), the coefficient of \( y \), to get \( 4 \), then square it to find \( 16 \).
• Rewrite \( y^2 + 8y \) as \((y+4)^2 - 16 \).

Substitute these expressions back into the equation to get it closer to the standard form. This manipulation is key to transforming complex quadratics into a recognizable form that can easily inform us of the features of the graph.

Center and Radius

Understanding the center and radius of a circle is fundamental when working with circle equations and graphing. From the standard form of a circle's equation \((x-h)^2 + (y-k)^2 = r^2\), the center is \((h, k)\) and the radius is \(r\).After completing the square and simplifying the given equation \( x^2 + y^2 - 6x + 8y + 18 = 0 \) into the standard form \( (x-3)^2 + (y+4)^2 = 7 \), we can directly read the center and radius:

• Center: \((3, -4)\) This indicates the circle's central point in the coordinate plane.
• Radius: \( \sqrt{7} \) The radius represents the distance from the center to any point on the circle.

Knowing these components is vital for sketching the graph accurately. The center reveals the circle's location, while the radius determines its size. With these, you can confidently draw the circle with an understanding of its position and scale.