Problem 60 - Intermediate Algebra

Question

Write each equation in standard form, if it is not already so, and graph it. The problems include equations that describe circles, parabolas, ellipses, and hyperbolas. $$ x^{2}+y^{2}+8 x-2 y-8=0 $$

Step-by-Step Solution

Answer

The equation \((x+4)^2 + (y-1)^2 = 25\) describes a circle with center \((-4, 1)\) and radius 5.

Step 1: Simplify and Rearrange Terms

First, identify the type of equation. The terms \(x^2\) and \(y^2\) with positive coefficients suggest a circle. Rearrange the equation to group \(x\) and \(y\) terms: \[x^2 + 8x + y^2 - 2y = 8\].

Step 2: Complete the Square for x

To find the center and radius, complete the square. Starting with \(x\)-terms: \(x^2 + 8x\). Take half of the coefficient of \(x\) (which is 8), square it \((4^2 = 16)\), and add inside the equation: \[(x^2 + 8x + 16) - 16\].

Step 3: Complete the Square for y

Now work on \(y\)-terms: \(y^2 - 2y\). Half of the coefficient of \(y\) is 1, squared it equals 1. Modify the equation: \[(y^2 - 2y + 1) - 1\].

Step 4: Rewrite the Equation

Combine completed squares and adjust right side to maintain balance: \[(x+4)^2 - 16 + (y-1)^2 - 1 = 8\]. Simplify: \[(x+4)^2 + (y-1)^2 = 25\].

Step 5: Identify Center and Radius of Circle

The equation \((x+4)^2 + (y-1)^2 = 25\) is standard form of a circle \((x-h)^2 + (y-k)^2 = r^2\). Here, center \((h, k)\) is \((-4, 1)\) and radius \(r\) is \(\sqrt{25} = 5\).

Step 6: Graph the Circle

Plot the center at \((-4, 1)\) on the coordinate plane. From the center, measure 5 units in all directions (up, down, left, right) to outline the circle. Sketch the circle passing through these points.

Key Concepts

Equation of a Circle

Understanding the equation of a circle is crucial when working with circles in algebra. The standard form of the equation of a circle is given by \((x-h)^2 + (y-k)^2 = r^2\), where:

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

This equation reflects the distance relationship between any point \((x,y)\) on the circle and its center. In the exercise, the circle's equation was derived from the general equation by completion of the square, transforming \(x^2 + y^2 + 8x - 2y - 8 = 0\) into the form where the center and radius can easily be identified. This helps not only in graphing the circle but also in understanding its key geometric properties.

Completing the Square

Completing the square is a valuable technique in algebra used to rewrite quadratic expressions in a more useful form. This process lets us easily find the vertex, center, or focus of quadratic-related graphs like circles, parabolas, and more. When applied to the terms in a circle's equation, it can help transform the equation into its standard form.In the step-by-step solution, completing the square was used twice:

• For the \(x\) terms: Start with \(x^2 + 8x\), take half of the coefficient of \(x\), square it (producing 16), and form \((x+4)^2\) by adjusting the equation.
• For the \(y\) terms: Start with \(y^2 - 2y\), take half of the coefficient of \(y\), square it (producing 1), and form \((y-1)^2\) by adjusting the equation.

Completing the square "completes" each perfect square trinomial, making it straightforward to identify the standard form and subsequently the graph's center and radius.

Graphing Circles

Graphing a circle involves placing its center on the Cartesian plane and marking points at a distance equal to the radius, forming a full circular path.With the equation \((x+4)^2 + (y-1)^2 = 25\), the identified center at \((-4, 1)\) makes plotting easier. Start by:

• Plotting the center point \((-4, 1)\) on the graph.
• Measuring out the radius of 5 units around the center. This should be done in all four cardinal directions: up, down, left, and right.

These points serve as guides to draw the perimeter of the circle. Remember, every point on this circle is exactly 5 units from the center, ensuring accuracy in drawing the circle. This step is essential in visualizing and understanding the geometric and spatial aspects of algebraic problems.