Problem 62 - Precalculus with Limits

Question

In Exercises 57-72, classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. \(x^2+y^2-4x+6y-3=0\)

Step-by-Step Solution

Answer

The graph of the given equation represents a circle.

Step 1: Rearrange the given equation

Rearrange the equation by completing the square in x and y. This leads to the following equation: \((x-2)^2+ (y+3)^2=16\)

Step 2: Analyzing the standard form

This equation is now in the standard form of a circle \((x-a)^2 + (y-b)^2 = r^2\), where 'a' and 'b' are the coordinates of the center and 'r' is the radius of the circle.

Step 3: Classify the graph

Since the equation is in the standard form of a circle and there is no variation in coefficients under the square terms and the sign between them is positive, this implies that the equation represents a circle

Key Concepts

Equation of a Circle

A circle in mathematics is defined as the set of all points in a plane that are at a fixed distance from a given point, called the center. To express this in the form of an equation, we use the formula \[(x-a)^2 + (y-b)^2 = r^2\]where:

• \(a\) and \(b\) represent the x and y coordinates of the circle's center, respectively.
• \(r\) is the radius, or the fixed distance from the center to any point on the circle.

In the exercise, the equation \((x-2)^2 + (y+3)^2 = 16\) describes a circle where the center is located at \((2, -3)\), and the radius is \(\sqrt{16} = 4\). Understanding this fundamental equation helps in identifying how a circle maintains its form and properties in a coordinate plane.

Completing the Square

Completing the square is a technique used to simplify quadratic equations into a form that is easier to work with, particularly when identifying the geometric shape being described. This method involves rearranging an equation to create a perfect square trinomial.

For the given problem:

• The original equation is \(x^2 + y^2 - 4x + 6y - 3 = 0\).
• To complete the square for \(x\), you rearrange \(x^2 - 4x\) to \((x-2)^2 - 4\).
• For \(y\), rearrange \(y^2 + 6y\) to \((y+3)^2 - 9\).
• Combine these to rewrite the equation as \((x-2)^2 + (y+3)^2 = 16\).

By completing the square, you transform the given equation into one that clearly shows its geometrical properties, revealing that it’s a circle with a specific center and radius.

Standard Form of Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. These include circles, ellipses, parabolas, and hyperbolas. Each has a specific standard form, used to easily identify and classify them based on their equation.

- **Circle**: \((x-a)^2 + (y-b)^2 = r^2\)- **Ellipse**: \(\frac{(x-a)^2}{b^2} + \frac{(y-b)^2}{c^2} = 1\)- **Parabola**: \(y = ax^2 + bx + c\)- **Hyperbola**: \(\frac{(x-a)^2}{b^2} - \frac{(y-b)^2}{c^2} = 1\)

Recognizing the standard form enables you to classify the graph of an equation swiftly. In the example, the equation is rearranged into the circle’s standard form, identifying it conclusively as a circle. Understanding these forms is essential in analyzing and solving various geometric problems.