Question

Name the conic or limiting form represented by the given equation. Usually you will need to use the process of completing the square (see Examples \(3-5) .\) x^{2}+y^{2}+6 x-2 y+6=0

Short answer

The given equation represents a circle.

Step-by-step solution

Group and Reorganize Terms

Start by reorganizing the terms in the given equation \(x^2 + y^2 + 6x - 2y + 6 = 0\). Group the \(x\)-terms together and \(y\)-terms together: \((x^2 + 6x) + (y^2 - 2y) = -6\). The constant term is moved to the other side of the equation.

Complete the Square for x-terms

Take the \(x\)-terms \(x^2 + 6x\). To complete the square, add and subtract \((\frac{6}{2})^2 = 9\). The expression becomes \((x^2 + 6x + 9 - 9) = (x + 3)^2 - 9\). Update the equation to: \((x + 3)^2 - 9 + (y^2 - 2y) = -6\).

Complete the Square for y-terms

Take the \(y\)-terms \(y^2 - 2y\). To complete the square, add and subtract \((\frac{-2}{2})^2 = 1\). The expression becomes \((y^2 - 2y + 1 - 1) = (y - 1)^2 - 1\). Update the equation to: \((x + 3)^2 - 9 + (y - 1)^2 - 1 = -6\).

Simplify the Equation

Substitute back the completed square forms into the equation: \((x + 3)^2 - 9 + (y - 1)^2 - 1 = -6\). Combine like terms: \((x + 3)^2 + (y - 1)^2 - 10 = -6\). Simplify to: \((x + 3)^2 + (y - 1)^2 = 4\).

Identify the Conic Section

The equation \((x + 3)^2 + (y - 1)^2 = 4\) is in the standard form of a circle \((x-h)^2 + (y-k)^2 = r^2\), where \(h = -3\), \(k = 1\), and \(r = 2\). Thus, the conic represented is a circle.

Key concepts

Completing the Square

Completing the square is a useful algebraic technique that helps to transform quadratic expressions into a perfect square trinomial. This method is particularly helpful when you want to rewrite quadratic equations in a way that makes them easier to solve or analyze.
To complete the square for a quadratic expression like \( ax^2 + bx + c \), follow these steps:

• Ensure the coefficient of the quadratic term (\( ax^2 \)) is 1. If not, factor out the coefficient from both the \( x^2 \) and \( x \) terms.
• Take the coefficient of the \( x \) term, divide it by 2, and square the result.
• Add and subtract this square inside the expression, effectively reshaping the quadratic into a perfect square trinomial.
• Factor the perfect square trinomial into the format \((x+d)^2\), where \(d\) is half of the \(x\) term's original coefficient.

In the example provided, the \(x\) terms \(x^2 + 6x\) were transformed into \((x + 3)^2 - 9\) by adding and subtracting 9. This transformation enables the equation to be easier to manage when determining the conic section classification.

Circle Equation

The circle equation in its most recognizable form is derived from completing the square, as seen in the conversion of the original equation in the exercise to a circle's form. The typical equation of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \( (h, k) \) is the center and \( r \) is the radius.
When an equation involving \(x\) and \(y\) is given, the goal is to rewrite it into this standard circle form. By achieving the circle equation, you can seamlessly identify the circle's key components:

• The center \((h, k)\), coming from the values that maintain the balance of the squares when the equation equals the radius squared.
• The radius \( r \), which is the square root of the constant on the other side of the equation from the squared terms.

Transforming the original equation \((x + 3)^2 + (y - 1)^2 = 4\) revealed that the equation describes a circle centered at \((-3, 1)\) with a radius of 2. This is an intuitive equation revealing both position and size of the circle.

Standard Form of a Circle

The standard form of a circle's equation is convenient for both identification and calculation. This can be seen as the goal when applying the completing the square method to the given equation of any quadratic relation involving two variables.
The equation of a circle is structured as \((x-h)^2 + (y-k)^2 = r^2\). Here's why this form is especially useful:

• The terms \((x-h)^2\) and \((y-k)^2\) quickly point to the center of the circle \((h, k)\), making it easy to understand where the circle sits in a coordinate plane.
• The equation solves easily for both graphical and algebraic approaches in geometrical problems.
• It provides an immediate visual representation through the center and radius, allowing for simpler interpretations and calculations in a wider range of mathematical problems.

In the exercise solution, the original complex quadratic was rearranged into exactly this standard form, demonstrating how mathematical techniques can simplify problem-solving to recognize geometric shapes and their properties effortlessly.