Problem 62 - College Algebra Essentials

Question

complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}+y^{2}+x+y-\frac{1}{2}=0 $$

Step-by-Step Solution

Answer

The standard form of the equation is \((x+\frac{1}{2})^{2} + (y+\frac{1}{2})^{2} =1\), the center of the circle is \(-\frac{1}{2}, -\frac{1}{2}\) and the radius is 1.

Step 1: Group the terms

Rearrange the equation by grouping the terms: \[ (x^{2}+x)+(y^{2}+y)=\frac{1}{2} \]

Step 2: Completing square for each group

To continue, we will need to form perfect squares.For (x^2 + x), the formula of completing the square comes in handy: (x + a)^2 = x^2 + 2ax + a^2, with a being half of the coefficient of x which is \(\frac{1}{2}\). Completing the square of (x^2 + x), we get \((x+\frac{1}{2})^2 - \frac{1}{4}\).Similarly, for (y^2 + y), the same method gives \((y + \frac{1}{2})^2 - \frac{1}{4}\). Substituting these results back in the equation, we get: \((x +\frac{1}{2})^{2} - \frac{1}{4} + (y+\frac{1}{2})^{2} - \frac{1}{4} =\frac{1}{2}\)

Step 3: Simplification and Resulting Standard Form

Simplify the equation by adding similar terms which results in: \((x+\frac{1}{2})^{2} + (y+\frac{1}{2})^{2} =1\). This is the standard form of the equation for a circle with center at \(-a, -b\) and radius \(r\).

Step 4: Determine the Center and the Radius of the Circle

From the standard form, we can infer that the center of the circle is \(-\frac{1}{2}, -\frac{1}{2}\) and the radius of the circle is the square root of 1 which is 1.

Step 5: Graph the Circle

On a coordinate plane, plot the center of the circle at \(-\frac{1}{2}, -\frac{1}{2}\). Then, draw a circle with a radius of 1 unit centered at this point.

Key Concepts

Standard Form of a Circle Equation

When dealing with circles in algebra, it’s helpful to know the standard form of a circle equation. This form looks like this: \[(x - h)^2 + (y - k)^2 = r^2\] In this equation,

• \((h, k)\) represents the center of the circle.
• \(r\) is the radius.
• \((x, y)\) represents any point on the circle’s edge.

By having the equation in this format, identifying the circle's center and radius is straightforward and visually intuitive. Notice that each squared term represents a translation in either the x or y direction, helping define the circle's specific location and size on a graph. Completing the square is a crucial step to take a general quadratic circle equation and convert it into this standard form, making analysis and graphing much simpler.

Center of a Circle

The center of a circle, indicated by the coordinates \((h, k)\) in the standard form equation, is the point from which every point on the circle is equidistant. It acts as a focal point for measuring the circle's radius.
In the example from the exercise, the center is calculated as \(-\frac{1}{2}, -\frac{1}{2}\). This is derived from reformulating the original equation into the standard circle equation format, and adjusting signs during the process of completing the square.

• The sign in the equation \((x + \frac{1}{2})^2 + (y + \frac{1}{2})^2\) tells us that the center coordinates are actually the opposites, converting to negative when interpreted from the original form.
• This transformation happens because the general form \((x-a)^2\) suggests a move to the right for positive \(a\) or to the left for negative \(a\).

Understanding how to extract the center from the equation will aid greatly in plotting and assessing circles better on a graph.

Radius of a Circle

The radius of a circle is the length from the center to any point on the perimeter of the circle. In the standard form equation \[(x - h)^2 + (y - k)^2 = r^2\], \(r\) represents the radius.
For the exercise at hand, we derived \(r^2 = 1\), meaning\(r = \sqrt{1} = 1\). This indicates that the circle extends equally one unit in every direction from its center.

• The radius is a non-negative value that directly shows how large or small the circle is.
• In real-world applications, knowing the radius can help determine the circle's circumference (via \(2\pi r\)) or its area (via \(\pi r^2\)).

Having a firm grasp on how to determine the radius from the standard circle equation is imperative for effective problem-solving in many mathematical tasks.