Problem 53 - Precalculus Essentials

Question

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+6 x+2 y+6=0$$

Step-by-Step Solution

Answer

The center of the circle is (-3, -1) and the radius is 2.

Step 1: Grouping

Group the x and y terms together. This gives \(x^{2}+6x + y^{2}+2y + 6 = 0\)

Step 2: Completing the square for 'x' terms

To complete the square, take half of the coefficient of the x term, square it and add it to both sides of the equation. The coefficient of x is 6, and half of 6 is 3. Squaring 3 gives 9. Add 9 to both sides to get \((x^{2}+6x+9) + y^{2}+2y + 6 = 9\)

Step 3: Completing the square for 'y' terms

Repeat the process for y. The coefficient of y is 2, so half of 2 is 1. Squaring 1 equals 1. Add 1 to both sides to get \((x^{2}+6x+9) + (y^{2}+2y+1) + 6 = 9 + 1\)

Step 4: Writing in standard form

Rewrite the equation so that it is in the standard form for the equation of a circle. You can also simplify the right side of the equation. The equation now becomes \((x+3)^{2}+(y+1)^{2}=4\)

Step 5: Finding center and radius

The standard form of the equation of a circle is \((x - a)^{2}+(y - b)^{2}=r^{2}\) where the centre is (a, b) and the radius is r. In our equation, we have the form \((x + 3)^{2}+(y + 1)^{2}= 4\). Therefore the center of the circle is (-3, -1) and the radius is the square root of 4, which is 2.

Step 6: Graphing the circle

Graph the circle using the center and radius. The center at (-3, -1) will be the point from which a circle of radius 2 will be drawn.

Key Concepts

Equation of a Circle

The equation of a circle in mathematics can be represented in various forms, but the most commonly used is known as the standard form of a circle equation. This form is beneficial because it provides clear and straightforward information about the circle's center and radius. The standard form is expressed as: \[(x - a)^{2} + (y - b)^{2} = r^{2}\] where:

• \(a\) and \(b\) are the x and y coordinates of the circle's center, respectively.
• \(r\) is the circle's radius.

When we are given an equation like \(x^{2} + y^{2} + 6x + 2y + 6 = 0\), it's not straightforward to see the properties of the circle. This is because the equation isn't in its standard form. To transform it into the standard form, we need to use the technique of completing the square. This method allows us to rewrite the equation, so it becomes possible to identify the center and radius of the circle easily.

Center and Radius of Circle

Determining the center and radius of a circle starts with converting the given equation into its standard form. As previously mentioned, the standard form is: \[(x - a)^{2} + (y - b)^{2} = r^{2}\]To find the center and radius from an equation such as \((x + 3)^{2} + (y + 1)^{2} = 4\), follow these steps:

• Recognize that the expression \((x + 3)^{2}\) indicates a horizontal shift in the x-direction by \(-3\). This means our \(a\) is \(-3\).
• Similarly, \((y + 1)^{2}\) suggests a vertical shift by \(-1\), implying \(b\) is \(-1\).
• The number on the right side of the equation, \(4\), represents \(r^2\). To find the radius \(r\), simply take the square root of \(4\), which is \(2\).

Therefore, in this equation, the circle has a center at \((-3, -1)\) and a radius of \(2\). This clear and concise representation is what makes the standard form advantageous for solving problems related to circles.

Standard Form

The standard form of a circle is essential in understanding and graphing its equation. After completing the square for both the x and y terms in an equation like \(x^2 + 6x + y^2 + 2y + 6 = 0\), you'll rewrite it in standard form. Here's how to do it:

• First, separate the x terms and y terms, as in \(x^2 + 6x\) and \(y^2 + 2y\). Then complete the square for each group.
• For the x terms: Take half of the coefficient of \(x\) (which is \(6\)), get \(3\), square it to get \(9\), then add \(9\).
• For the y terms: Take half of the coefficient of \(y\) (which is \(2\)), get \(1\), square it to get \(1\), and add \(1\).

This process results in an equation such as \((x+3)^{2} + (y+1)^{2} = 4\), which represents the standard form. By comparing this form to \((x-a)^2 + (y-b)^2 = r^2\), you obtain critical information about the circle's center and radius. This convenience is why the standard form is used so widely in equations involving circles.