Problem 59 - 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. $$ x^{2}+y^{2}+4 x-6 y-23=0 $$

Step-by-Step Solution

Answer

Equation in standard form: \((x+2)^{2} + (y-3)^{2} = 36\); Center: (-2,3); Radius: 6.

Step 1: Identifying the Type of Equation

The given equation is \(x^{2}+y^{2}+4x-6y-23=0\). Since both \(x\) and \(y\) are squared and their coefficients are equal, the equation represents a circle.

Step 2: Rewriting the Equation

We need to rewrite the equation in the standard form of a circle, which is \((x-h)^{2}+(y-k)^{2}=r^{2}\). Our goal will be to complete the square for both the \(x\) and \(y\) terms.

Step 3: Completing the Square for x

For the \(x\) terms, the expression is \(x^{2}+4x\). Take half of the \(4\), which is \(2\), square it to get \(4\), and add \(4\) to both sides of the equation: \(x^{2} + 4x + 4\).

Step 4: Completing the Square for y

For the \(y\) terms, the expression is \(y^{2}-6y\). Take half of \(-6\), which is \(-3\), square it to get \(9\), and add \(9\) to both sides of the equation: \(y^{2} - 6y + 9\).

Step 5: Rewrite the Equation with Completed Squares

Adding these squares to the left side changes the original equation to: \((x+2)^{2} + (y-3)^{2} = 23 + 4 + 9\). Simplifying yields: \((x+2)^{2} + (y-3)^{2} = 36\).

Step 6: Graphing the Circle

The equation \((x+2)^{2} + (y-3)^{2} = 36\) is in standard form. The center of the circle is \((-2, 3)\), and the radius is \(6\) (since \(r^{2}=36\), so \(r=6\)). Plot the center on the graph and draw a circle with radius 6.

Key Concepts

Completing the Square

Completing the square is a method used in algebra to transform a quadratic expression into a perfect square trinomial. This process is particularly useful when dealing with equations of circles, as it helps to rewrite those equations in standard form.

To complete the square for a term like \(x^2 + bx\), you follow these steps:

• Identify the coefficient of \(x\), which is \(b\).
• Take half of \(b\), resulting in \(\frac{b}{2}\).
• Square this, \(\left(\frac{b}{2}\right)^2\), and add it to both sides of the equation to maintain equality.

The same process applies to the \(y\) terms.

This method is essential for converting equations like \(x^2 + y^2 + 4x - 6y - 23 = 0\) into a form that is easily recognizable as a circle. By completing the square, you identify the transformations needed for graphing and interpreting the equation in terms of a circle’s position and size. Completing the square simplifies the equation and exposes features like the circle's center and radius.

Standard Form of a Circle

The standard form of a circle's equation is given by \((x-h)^2 + (y-k)^2 = r^2\). In this equation, \((h, k)\) represents the center of the circle, and \(r\) is the radius. This formulation makes it straightforward to understand the circle's properties.

By completing the square on a general quadratic equation that includes both \(x^2\) and \(y^2\) terms with linear terms, you can identify \(h\), \(k\), and \(r\). For example, the equation \((x+2)^2 + (y-3)^2 = 36\) has:

• Center at \((-2, 3)\)
• Radius of 6, since \(r^2 = 36\)

This form is particularly useful when analyzing and graphing circles, as it directly reveals the circle’s center point and its radius. Understanding the standard form enables you to easily switch from equation to visual representation on a graph.

Graphing Circles

Graphing circles involves plotting a perfect round shape on a coordinate plane, based on its standard form equation \((x-h)^2 + (y-k)^2 = r^2\). The graph of a circle is defined by its center \((h, k)\) and radius \(r\).

Here are key steps to follow when graphing:

• Identify the center \((h, k)\) from the equation.
• Determine the radius \(r\) by taking the square root of \(r^2\).
• Plot the center point on the graph.
• Use the radius to mark key points around the center in each direction – up, down, left, and right – on the coordinate plane.
• Draw a smooth curve connecting these points, ensuring it's equidistant from the center.

Graphing is about understanding the relationship between algebraic expressions and their geometric representations. A well-graphed circle allows you to visually confirm properties such as symmetry and location that are represented algebraically in the equation. Proper graphing reinforces the link between equations and their visual counterparts.