Problem 39 - Intermediate Algebra

Question

\(x^{2}+y^{2}+6 x-2 y+6=0\)

Step-by-Step Solution

Answer

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

Step 1: Rearrange the Equation

First, we'll rearrange the given equation by grouping the terms with the same variables together. The equation is originally \( x^2 + y^2 + 6x - 2y + 6 = 0 \). Group similar terms together to get:\[ x^2 + 6x + y^2 - 2y = -6 \]

Step 2: Complete the Square for \(x\)-Terms

Focus on the \(x\)-terms: \(x^2 + 6x\). To complete the square, take half of the coefficient of \(x\), which is 6, divide it by 2 to get 3, and then square it to obtain 9. Add and subtract 9:\[ (x^2 + 6x + 9) - 9 \]

Step 3: Complete the Square for \(y\)-Terms

Now, work on the \(y\)-terms: \(y^2 - 2y\). Again, take half of -2 which is -1, square it to get 1. Add and subtract 1:\[ (y^2 - 2y + 1) - 1 \]

Step 4: Rewrite the Equation with Perfect Squares

The equation has now been transformed by completing the square for both variables:\[ (x + 3)^2 - 9 + (y - 1)^2 - 1 = -6 \]This simplifies to:\[ (x + 3)^2 + (y - 1)^2 = 4 \]

Step 5: Recognize the Equation of a Circle

The equation \( (x + 3)^2 + (y - 1)^2 = 4 \) resembles the standard form of a circle equation, \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center and \(r\) is the radius. Here, the center is \((-3, 1)\) and the radius is \(2\).

Key Concepts

Equation of a Circle

The equation of a circle is an essential concept in geometry. It is expressed in the standard form as \( (x - h)^2 + (y - k)^2 = r^2 \), where

• \((h, k)\) represents the coordinates of the circle’s center.
• \(r\) stands for the radius of the circle.

This equation helps determine the position and size of a circle in the coordinate plane.

In the given problem, we transformed the equation \( x^2 + y^2 + 6x - 2y + 6 = 0 \) into the circle's standard form. By completing the square, which we'll cover next, we determined the center and radius. The final form \( (x + 3)^2 + (y - 1)^2 = 4 \) allowed us to identify the center as \((-3, 1)\) and the radius as \(2\). This process signifies rearranging a quadratic equation to match the circle's equation format for better understanding.

Quadratic Equations

Quadratic equations appear frequently in algebra and can generally be expressed in the standard form: \( ax^2 + bx + c = 0 \). Here,

• \(a, b\), and \(c\) are constant coefficients,
• \(x\) is the variable.

These equations are pivotal as they represent parabolas in the xy-plane when graphed.

For the provided exercise, the quadratic nature was evident in the terms \(x^2\) and \(y^2\). Understanding quadratic equations is integral to manipulating and completing the square steps used in transforming the original equation into the one describing a circle.

Algebraic Manipulation

Algebraic manipulation is a technique to rearrange or simplify expressions. It's a vital skill in solving equations and making them more understandable or usable in further calculations.

• Combine like terms.
• Use operations like addition, subtraction, multiplication, or division consistently.
• Transform equations to achieve useful forms, like factoring or completing the square.

These are all part of algebraic manipulation strategies.

In the context of our exercise, algebraic manipulation was used to "complete the square." This involved adjusting terms to form perfect squares, turning the equation \((x^2 + 6x)\) into \((x+3)^2\) and \((y^2 - 2y)\) into \((y-1)^2\). This manipulation made identifying the circle's equation easier, showcasing the power and flexibility of algebraic manipulation in solving and simplifying complex problems.