Problem 58 - Precalculus

Question

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$x^{2}+y^{2}-4 x+6 y-3=0$$

Step-by-Step Solution

Answer

The graph of the equation \(x^{2}+y^{2}-4 x+6 y-3=0\) represents a circle.

Step 1: Group x terms and y terms

Rearrange the equation by grouping the x terms and y terms together. The equation becomes: \(x^{2}-4x+y^{2}+6y-3=0\)

Step 2: Complete the square for x and y terms

The next step is to complete the square, add and subtract the square of half the coefficient of x and y inside the brackets respectively.\nThe equation can be rewritten as: \((x^{2}-4x+4)+(y^{2}+6y+9)-3-4-9=0\).\nSo, the equation simplifies to: \[(x-2)^{2}+(y+3)^{2}=6\]

Step 3: Identify the conic section

By looking at the equation in Step 2, it has a square term for both x and y, so it can be a circle, ellipse or a hyperbola. Here, the coefficients of both squares are the same, meaning the graph represents a circle

Key Concepts

Circle Equation

Understanding the circle equation is essential for visualizing and solving problems related to circular graphs. A circle equation typically takes the form \( (x-h)^2 + (y-k)^2 = r^2 \) where \( (h, k) \) is the center of the circle and \( r \) is its radius. Starting from the general equation of a circle, we can manipulate any quadratic equation to match this form using certain algebraic techniques, allowing us to identify circles among other conic sections.

For the given equation \( x^2 + y^2 - 4x + 6y - 3 = 0 \) from our exercise, our goal is to rearrange and recognize it as a circle equation. By completing the square (which we’ll address in more detail shortly), we transform the given quadratic into the standard circle form. The crucial aspect to note is that, in a circle equation, the coefficients of \( x^2 \) and \( y^2 \) are equal and the square terms have the same sign. This is a clear signpost that we are dealing with a circle when classifying conic sections.

Completing the Square

Completing the square is a powerful method used in algebra to transform a quadratic expression into a perfect square trinomial. For equations involving conic sections, this technique is instrumental in morphing them into a recognizable standard form. Here’s the crux of the process: take the coefficient of the linear term (that's the term with just \( x \) or \( y \) in it), halve it, square it, and then add and subtract this squared value to the expression.

The exercise provided demonstrates this process well. The equation \( x^2 - 4x + y^2 + 6y - 3 = 0 \) needs to be rewritten. By adding and subtracting the square of half the coefficients of \( x \) and \( y \) to the equation, we turn the x and y terms into perfect square trinomials, which simplifies the classification of the graph. Through completing the square, we move from a mere quadratic expression to the specific form that reveals the nature of the conic section, in this case, a circle equation.

Graph Classification

Graph classification is crucial in identifying the type of conic section represented by a given equation. Conic sections include circles, ellipses, parabolas, and hyperbolas, each having its distinct equation form. Here’s a quick guide to distinguish between them:

• Circles have equal coefficients for \( x^2 \) and \( y^2 \) with opposite signed constants.
• Ellipses also have equal coefficients, but like circles, no cross-product term (like \( xy \) exists).
• Parabolas feature only one squared term, either \( x^2 \) or \( y^2 \) but not both.
• Hyperbolas have squared terms of \( x \) and \( y \) with opposite signs (one positive and one negative).

After completing the square for our equation, the result is \[ (x-2)^2 + (y+3)^2 = 16 \], a standard form circle equation. It’s clear that both squared terms have the same coefficient (in this case, the implied coefficient is 1), leading us to classify the graph as a circle. By recognizing these patterns, we set a foundation for solving conic section equations and understanding their respective graphs.