Problem 45 - A Graphical Approach to Precalculus with Limits

Question

Decide whether each equation has a circle as its graph. If it does, give the center and radius. $$x^{2}+6 x+y^{2}+8 y+9=0$$

Step-by-Step Solution

Answer

The equation is a circle with center (-3, -4) and radius 4.

Step 1: Rewrite the Equation

The given equation is \(x^{2}+6x+y^{2}+8y+9=0\). First, move the constant term to the right side of the equation. This gives us \(x^{2}+6x+y^{2}+8y = -9\).

Step 2: Complete the Square

To complete the square, we'll handle the \(x\) and \(y\) terms separately. For the \(x\) terms: 1. Take the coefficient of \(x\), which is 6, divide it by 2 to get 3, and square it to get 9. 2. Add and subtract 9 to the equation for balancing: \(x^2 + 6x + 9 - 9\).For the \(y\) terms:1. Take the coefficient of \(y\), which is 8, divide it by 2 to get 4, and square it to get 16.2. Add and subtract 16 to the equation for balancing: \(y^2 + 8y + 16 - 16\).The equation is now: \((x^2 + 6x + 9) + (y^2 + 8y + 16) = -9 + 9 + 16\).

Step 3: Simplify and Rearrange

Now, rewrite the equation using the complete squares:\((x + 3)^2 + (y + 4)^2 = 16\).Notice that the equation can now be compared with the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), where \((h,k)\) is the center and \(r\) the radius.

Step 4: Identify the Circle's Center and Radius

In the equation \((x + 3)^2 + (y + 4)^2 = 16\), the center of the circle is \((-3, -4)\) and the radius is the square root of 16, which is 4.

Key Concepts

Completing the Square

Completing the square is a powerful method used in algebra to simplify quadratic equations. This process is essential when rewriting circle equations in standard form. Given the equation \(x^2 + 6x + y^2 + 8y + 9 = 0\), the first step is to move the constant to the other side, resulting in \(x^2 + 6x + y^2 + 8y = -9\).
To complete the square for the \(x\) terms:

• Take the coefficient of \(x\), which is 6.
• Divide by 2 to get 3.
• Square this result to get 9.
• Add and subtract 9 to maintain the equation's balance.

For the \(y\) terms:

• Take the coefficient of \(y\), which is 8.
• Divide by 2 to get 4.
• Square this result to get 16.
• Add and subtract 16 as well.

Now, the equation becomes \((x + 3)^2 + (y + 4)^2 = 16\). By completing the square, we've converted the equation into one that is easy to interpret as a circle's equation.

Center of a Circle

The center of a circle provides the fixed point from which all points on the circle have the same distance, known as the radius. The standard form of a circle equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center. Our completed equation \((x + 3)^2 + (y + 4)^2 = 16\) reveals the center after careful observation.
Notice the terms inside the parenthesis:

• In \((x + 3)^2\), the \(+3\) indicates a horizontal shift left by 3 units, giving \(h = -3\).
• In \((y + 4)^2\), the \(+4\) indicates a vertical shift downward by 4 units, giving \(k = -4\).

Thus, the center of the circle is \((-3, -4)\). Knowing the center is crucial for understanding the circle's position on a coordinate plane.

Radius of a Circle

The radius of a circle is the constant distance from the center to any point on the circle. It is derived from the equation \((x-h)^2 + (y-k)^2 = r^2\), where \(r\) is the radius. In our equation **\((x + 3)^2 + (y + 4)^2 = 16\)**, the circle is expressed in standard form, making it easy to identify the radius.
The right side of the equation, \(16\), represents \(r^2\), the square of the radius.

• To find \(r\), take the square root of 16.
• \(r = \sqrt{16} = 4\).

Therefore, the circle's radius is 4 units. The radius is instrumental in understanding the scale and size of the circle, as well as calculating the area and circumference in more advanced applications.