Problem 56 - A Graphical Approach to College Algebra

Question

Decide whether each equation has a circle as its graph. If it does, give the center and radius. $$9 x^{2}+9 y^{2}+54 y=-72$$

Step-by-Step Solution

Answer

Yes, it's a circle with center (0, -3) and radius 1.

Step 1: Identify Circle Equation Form

The general equation of a circle is given by \((x-h)^2 + (y-k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius. We need to rearrange and complete the square to match this format with what's given: \(9x^2 + 9y^2 + 54y = -72\).

Step 2: Rearrange the Equation

First, move all terms to one side to allow completing the square. Subtract each term on the right by adding 72: \[9x^2 + 9y^2 + 54y + 72 = 0.\]

Step 3: Factor out the Coefficient of the Quadratic Terms

Factor out 9 from the terms involving \(x\) and \(y\): \[9(x^2 + y^2 + 6y) = -72.\]

Step 4: Complete the Square for Y

Focus on completing the square for \(y\) in \(y^2 + 6y\). The step is to take half of the coefficient of \(y\), square it, and add/subtract inside the equation: \[y^2 + 6y = (y+3)^2 - 9.\]

Step 5: Substitute and Simplify

Substitute the completed square back into the equation: \[9(x^2 + (y+3)^2 - 9) = -72.\] Simplify: \[9(x^2 + (y+3)^2) - 81 = -72.\] Add 81 to both sides: \[9(x^2 + (y+3)^2) = 9.\]

Step 6: Solve for Circle Equation

Divide the entire equation by 9 to isolate \((x^2 + (y+3)^2)\): \[(x^2 + (y+3)^2) = 1.\] This equation represents a circle.

Step 7: Determine Center and Radius

Compare the equation \((x^2 + (y+3)^2) = 1\) to \((x-h)^2 + (y-k)^2 = r^2\). The center \((h, k)\) is \((0, -3)\) and the radius \(r\) is 1.

Key Concepts

Completing the Square

Completing the square is a technique used to rearrange a quadratic equation. This helps when you want to convert it into a format that is easier to work with, like the equation of a circle. Imagine you have a quadratic expression like \(y^2 + 6y\). To "complete the square," follow these steps:

• Take half of the coefficient of \(y\), which is 6, and divide it by 2 to get 3.
• Square the number you found: \(3^2 = 9\).
• Add and subtract this square within the equation: \(y^2 + 6y + 9 - 9\).
• Now, \(y^2 + 6y + 9\) becomes \((y + 3)^2\) when factored.

Using this method, you can change the original equation into a simpler one where the circle's properties are clear. Completing the square often transforms the equation into something like \((y+k)^2\), which is closer to the standard form of a circle equation.

Equation of a Circle

The equation of a circle is elegantly simple. It's written as \((x-h)^2 + (y-k)^2 = r^2\). In this form:

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

To identify whether an equation like \(9x^2 + 9y^2 + 54y + 72 = 0\) is a circle, you need to rearrange it into the circle's equation form. This involves several steps:1. Move all terms to one side to prepare for completing the square.2. Factor common coefficients out of quadratic terms (here, factor out 9).3. Complete the square for any variables (like \(y\) in this case).4. Rearrange so that \((x-h)^2\) and \((y-k)^2\) form the main part of the equation, equaling some constant \(r^2\). Once in this form, the equation clearly shows the attributes of the circle. This clear format is essential for determining the circle's properties.

Center and Radius of a Circle

Once your equation is in the form \((x-h)^2 + (y-k)^2 = r^2\), it's straightforward to identify the center and radius of the circle:

• The center is at point \((h, k)\), taken directly from the equation. In our example after completing the square and rearranging, it is at \((0, -3)\).
• The radius \(r\) is the square root of the number on the other side. If the equation is \((x)^2 + (y+3)^2 = 1\), then \(r = \sqrt{1} = 1\).

Understanding where these values come from helps visualize the circle on a coordinate plane. It's not just about solving for \(h\), \(k\), and \(r\), but comprehending how each part of the equation represents a real feature of the circle, expanding your understanding of geometry.