Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

Name the conic or limiting form represented by the given equation. Usually you will need to use the process of completing the square. $$ x^{2}+y^{2}-2 x+2 y+1=0 $$

Step-by-Step Solution

Answer

The equation represents a circle with center (1, -1) and radius 1.

Step 1: Group x and y Terms

The given equation is \(x^{2} + y^{2} - 2x + 2y + 1 = 0\). Start by grouping the \(x\) terms and the \(y\) terms together: \((x^{2} - 2x) + (y^{2} + 2y) + 1 = 0\).

Step 2: Complete the Square for x Terms

For the \(x\) terms, take the coefficient of \(x\), which is \(-2\), divide by 2 to get \(-1\), and then square it to get \(1\). Add and subtract this inside the parentheses: \((x^{2} - 2x + 1 - 1)\). After this step, it becomes \((x - 1)^{2} - 1\).

Step 3: Complete the Square for y Terms

For the \(y\) terms, take the coefficient of \(y\), which is \(2\), divide by 2 to get \(1\), and then square it to get \(1\). Add and subtract this inside the parentheses: \((y^{2} + 2y + 1 - 1)\). After this step, it becomes \((y + 1)^{2} - 1\).

Step 4: Rewrite the Equation in Completed Square Form

Substitute the completed square forms back into the equation: \((x - 1)^{2} - 1 + (y + 1)^{2} - 1 + 1 = 0\).

Step 5: Simplify the Equation

Combine the constant terms: \((x - 1)^{2} + (y + 1)^{2} - 1 = 0\) simplifies to \((x - 1)^{2} + (y + 1)^{2} = 1\).

Step 6: Identify the Conic Section

The equation \((x - 1)^{2} + (y + 1)^{2} = 1\) fits the standard form of a circle, \((x - h)^{2} + (y - k)^{2} = r^{2}\), where \((h, k) = (1, -1)\) and \(r = 1\).

Key Concepts

Completing the Square

Completing the square is a handy algebraic technique used to transform quadratic expressions or entire equations into a form that reveals important properties, such as the vertex of a parabola or the center and radius of a circle. Here's a simple breakdown of how it works:

• Start with a quadratic expression, like \(ax^2 + bx + c\). Focus on the \(x\) terms, \(ax^2 + bx\).
• If the coefficient \(a\) is not 1, first factor it out from the \(x^2\) and \(x\) terms.
• Take the coefficient of \(x\), divide it by 2, and then square the result. This gives you the number to add and subtract (inside the parentheses) to complete the square.
• Rewrite the quadratic expression using this squared number so it becomes a perfect square trinomial \((x - p)^2\), helping identify key features, such as the center of a circle.

For example, in the expression \(x^2 - 2x\), take the coefficient \(-2\), divide by 2 to get \(-1\), and square to get 1. So, \(x^2 - 2x\) becomes \((x - 1)^2 - 1\), revealing a perfect square minus a constant. This step is mirrored with the \(y\) terms too.

Circle Equation

A circle equation arises from the set of all points in a plane that are equidistant from a given point known as the center. In the context of algebra, this is expressed in the form \((x - h)^2 + (y - k)^2 = r^2\), where:

• \((h, k)\) are the coordinates of the circle's center.
• \(r\) is the radius of the circle, which represents the constant distance from the center to any point on the circle.

Understanding how to derive and recognize this structure in different forms of equations allows solutions like detecting circles from general quadratic equations possible.
In practical terms, after completing the square for both \(x\) and \(y\) terms, you often end with an equation ready to transform into this recognizable circle format. For instance, the process turns \(x^2 + y^2 - 2x + 2y + 1 = 0\) to \((x - 1)^2 + (y + 1)^2 = 1\), making it plain to see that it's a circle with a center at \((1, -1)\) and radius 1.

Standard Form of a Circle

The standard form of a circle equation is an algebraic representation that clearly shows the center and the radius of the circle. As a reminder, it looks like \((x - h)^2 + (y - k)^2 = r^2\), where:

• \((h, k)\) is the center of the circle. In our example, translating \((x - 1)^2 + (y + 1)^2 = 1\) shows \((h, k)\) as \((1, -1)\).
• \(r\), the radius, equals 1, derived from the number on the right side of the equation being \(r^2 = 1\), so \(r = \sqrt{1} = 1\).

This form is extremely useful in both theoretical and applied contexts because it provides immediate information about the circle's key features without additional algebra. Given any conic section equation resembling a quadratic form, genuinely converting and spotting this standard form fortifies its identity as a circle in geometry. Recognizing such a conversion aids in visualizing and graphing aspects or solutions to geometrical problems quickly and effectively.