Problem 69 - Algebra and Trigonometry

Question

\(65-72\) . Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}+x=0 $$

Step-by-Step Solution

Answer

The circle has center \((-\frac{1}{2}, 0)\) and radius \(\frac{1}{2}\).

Step 1: Identify the Equation Type

We are given the equation \(x^2 + y^2 + x = 0\). This equation is of the form \(x^2 + y^2 + ax + by + c = 0\), which can represent a circle. To confirm, let's rewrite it in the standard form of a circle.

Step 2: Complete the Square for 'x'

Rewrite the equation to group \(x\) terms together: \(x^2 + x + y^2 = 0\). Complete the square for \(x\) by taking half of the coefficient of \(x\) (which is \(1\)), squaring it to get \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\), and adjusting the equation:\[x^2 + x + \frac{1}{4} + y^2 = -\frac{1}{4}\].

Step 3: Simplify the Expression

The expression \(x^2 + x + \frac{1}{4}\) can be rewritten as \(\left(x + \frac{1}{2}\right)^2\), turning the equation into:\[\(\left(x + \frac{1}{2}\right)^2 + y^2 = \frac{1}{4}\)\].

Step 4: Identify the Center and Radius

The equation \(\left(x + \frac{1}{2}\right)^2 + y^2 = \frac{1}{4}\) is now in the standard form \((x - h)^2 + (y - k)^2 = r^2\).- The center \((h, k)\) of the circle is \((-\frac{1}{2}, 0)\).- The radius \(r\) is the square root of \(\frac{1}{4}\), which is \(\frac{1}{2}\).

Key Concepts

Completing the Square

When transforming an equation into the standard form of a circle, completing the square is a helpful method. It helps to simplify quadratic expressions. Let’s break down the steps to complete the square:

• First, focus on the terms involving one variable, such as the x-terms.
• Take the coefficient of the x term, which is the number in front of the x"). If the term is x"), consider the coefficient to be 1 here.
• Divide this coefficient by 2. For example, for the equation x^2 + x"): \(\frac{1}{2} = 0.5\).
• Next, square that result: \(0.5^2 = 0.25\).
• Add and subtract this square inside the equation to maintain balance.
• Rewrite the expression as a binomial square. For example, x^2 + x + 0.25"): becomes \(\left(x + 0.5\right)^2\).

This method allows you to rearrange expressions neatly, setting the stage to see equations in their standard forms, particularly for determining circle equations.

Center of a Circle

The center of a circle is crucial for understanding its position in the coordinate plane. For a circle equation in the standard form \((x - h)^2 + (y - k)^2 = r^2\), the center is denoted by the coordinates \((h, k)\). Here’s how to find it from the equation through a simple process:

• Look for transformations applied to the \(x\) and \(y\) terms.
• For \(\left(x + a\right)^2\): the transformation is \(x - (-a)\), indicating a horizontal shift by \(-a\).
• For \(\left(y + b\right)^2\): the transformation is \(y - (-b)\), showing a vertical shift by \(-b\).
• Remember: If you see \(x + 0.5\), this corresponds to \(x - (-0.5)\).

In the given example equation, the terms \((x + \frac{1}{2})^2 \) and \(y^2\) indicate shifts. Thus, the center is at \((-\frac{1}{2}, 0)\). Think of these as adjustments that 'move' the circle from the origin to its actual center point.

Radius of a Circle

The radius of a circle tells us how far away the boundary is from the center. In the circle equation \((x - h)^2 + (y - k)^2 = r^2\), the radius is represented by r. To find it:

• Identify the constant on the right side of the equation, which is \(r^2\).
• Take the square root of this constant to find the radius \((r)\).
• For instance, if the original equation is simplified to: \(\left(x + \frac{1}{2}\right)^2 + y^2 = \frac{1}{4}\), then \((r^2 = \frac{1}{4})\).
• The square root of \(\frac{1}{4}\) is \(\frac{1}{2}\).

In our example, the given equation confirms the radius is \(\frac{1}{2}\). The radius provides valuable information about the circle’s scale and how it spans the coordinate plane.