Problem 43 - Precalculus with Limits

Question

In Exercises 29-52, identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. \(x^2+y^2-2x+4y-31=0\)

Step-by-Step Solution

Answer

The given conic section is a Circle with center at (1, -2) and the radius is 6. Circle doesn't have vertices, foci, or eccentricity.

Step 1: Identify the Conic Section

Upon initial inspection, the equation \(x^2 + y^2 - 2x + 4y - 31 = 0\) can be seen as a general form of either a circle or an ellipse, since the coefficients of \(x^2\) and \(y^2\) are the same, but we need to transform it into its standard form to have a clearer picture of its form and properties. The transformation is done by completing the square.

Step 2: Transform the given equation into standard form

To complete the square for the equation, the coefficients of x and y terms are halved and squared to be added and subtracted within the equation.\nRewriting the equation by grouping \(x\) and \(y\) terms, we get: \(x^2 - 2x + y^2 + 4y = 31\). Now, add the square of half the coefficient of \(x\) and the square of half the coefficient of \(y\), i.e., \(1^2\) and \((2)^2\). \nDoing so we get the standard form of the equation as follows: \((x-1)^2 + (y+2)^2 = 36\).

Step 3: Identify the Conic and Its Properties

The resulting equation \((x-1)^2 + (y+2)^2 = 36\) is the standard form of a circle's equation with center at \((1, -2)\) and radius \(\sqrt{36} = 6\). Because it's a circle, it doesn't have vertices, foci, or eccentricity.

Step 4: Sketch the Graph

Draw the graph with center at point (1, -2) and a radius of 6 units. Mark the center. Draw a circle with the marked center as the origin and passing through points 6 units apart from the origin in all directions. Ensure that the circle touches x = -5, x = 7, y = -8, and y = 4.

Key Concepts

Circle

A **circle** is a simple geometric shape that is perfectly round, like the shape of a wheel or a dinner plate. It is defined as the set of all points in a plane that are equidistant from a fixed point called the center. The constant distance from the center to any point on the circle is called the radius.

In the problem described in the exercise, we have a circle in which we will identify specific characteristics like the center and the radius. Identifying these elements helps in understanding the geometry and in sketching the circle accurately on a graph. Circles are unique among conic sections because they have no vertices, foci, or eccentricity, which simplifies the study of their properties.

Standard Form

The **standard form** of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where

• \((h, k)\) is the center of the circle.
• \(r\) is the radius.

This form makes it easy to understand and graph the circle. By bringing the given equation into this format, we can directly read off the center and the radius.

In our exercise, we transformed the original equation by completing the square. After simplifying, we obtained \((x-1)^2 + (y+2)^2 = 36\), indicating a circle centered at \((1, -2)\) with a radius of \(6\). This conversion is crucial for visualizing and working with the circle's properties.

Completing the Square

**Completing the square** is a method used to transform a quadratic equation into a standard form. This technique is essential when working with conic sections to reveal their specific attributes.

Here's how it's done:

• Group the \(x\) and \(y\) terms separately.
• Take half of the linear coefficient of each variable, square it, and add it to both sides.
• For the term \(x^2 - 2x\), add \(1^2\) to complete the square.
• For the term \(y^2 + 4y\), add \(2^2\).

This process helps convert the equation \(x^2 + y^2 - 2x + 4y = 31\) into the standard form \((x-1)^2 + (y+2)^2 = 36\). Understanding this method is crucial for identifying and graphing circles and other conic sections.

Properties of Circle

The **properties of a circle** are essential for understanding and working with this conic section. A circle is characterized by its center and radius, which are directly visible in the standard form of its equation.

Here are key properties to remember:

• The **center** of the circle in the exercise is \((1, -2)\).
• The **radius** is \(6\), calculated as the square root of \(36\).
• The circle is perfectly symmetrical around its center.
• It encloses the maximum area with a minimum perimeter compared to any other conic section.

Knowing these properties helps in plotting the circle on a graph and understanding its geometric significance in various applications.