Problem 71 - Algebra and Trigonometry

Question

65–72 ? Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}-\frac{1}{2} x+\frac{1}{2} y=\frac{1}{8} $$

Step-by-Step Solution

Answer

The center is \( (\frac{1}{4}, -\frac{1}{4}) \) and the radius is \( \sqrt{\frac{3}{8}} \).

Step 1: Write the Standard Form of a Circle

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

Step 2: Rearrange Terms to Complete the Square

Rearrange the equation: \[ x^2 - \frac{1}{2}x + y^2 + \frac{1}{2}y = \frac{1}{8} \]We need to complete the square for both \(x\) and \(y\) terms.

Step 3: Complete the Square for x

For the \(x\) terms: \[ x^2 - \frac{1}{2}x \]To complete the square, take half of \(-\frac{1}{2}\), which is \(-\frac{1}{4}\), and square it to get \(\frac{1}{16}\).Add and subtract \(\frac{1}{16}\):\[ x^2 - \frac{1}{2}x + \left( \frac{1}{16} \right) - \frac{1}{16}\] becomes \[ (x - \frac{1}{4})^2 - \frac{1}{16}\].

Step 4: Complete the Square for y

For the \(y\) terms: \[ y^2 + \frac{1}{2}y \]To complete the square, take half of \(\frac{1}{2}\), which is \(\frac{1}{4}\), and square it to get \(\frac{1}{16}\).Add and subtract \(\frac{1}{16}\):\[ y^2 + \frac{1}{2}y + \left( \frac{1}{16} \right) - \frac{1}{16}\] becomes \[ (y + \frac{1}{4})^2 - \frac{1}{16}\].

Step 5: Substitute Completed Squares Back into Equation

Substitute the completed square terms back into the original equation:\[ (x - \frac{1}{4})^2 - \frac{1}{16} + (y + \frac{1}{4})^2 - \frac{1}{16} = \frac{1}{8} \]Combine and simplify:\[ (x - \frac{1}{4})^2 + (y + \frac{1}{4})^2 = \frac{1}{8} + \frac{1}{16} + \frac{1}{16} \]\[ (x - \frac{1}{4})^2 + (y + \frac{1}{4})^2 = \frac{4}{16} + \frac{2}{16} = \frac{6}{16} = \frac{3}{8} \].

Step 6: Identify the Center and Radius

From the equation \[ (x - \frac{1}{4})^2 + (y + \frac{1}{4})^2 = \frac{3}{8} \]The center is \( (\frac{1}{4}, -\frac{1}{4}) \) and the radius of the circle is \( \sqrt{\frac{3}{8}} \).

Key Concepts

Completing the Square

The concept of completing the square is essential when working with quadratic expressions, especially when converting a quadratic equation into a form that reveals specific characteristics, like the standard form of a circle's equation. To complete the square, you add and subtract the same value to transform a trinomial into a perfect square trinomial, meaning it can be expressed as \((x - h)^2\).

This is how it works: If you have a term of the form \(x^2 + bx\), to complete the square, take half of the coefficient of \(x\), square it, and then add and subtract this number to your expression. For example, in the equation:

• \(x^2 - \frac{1}{2}x\), the coefficient of \(x\) is \(-\frac{1}{2}\).
• Half of \(-\frac{1}{2}\) is \(-\frac{1}{4}\).
• Square \(-\frac{1}{4}\) to get \(\frac{1}{16}\).
• Add and subtract \(\frac{1}{16}\) to get a perfect square: \((x - \frac{1}{4})^2 - \frac{1}{16}\).

By repeating this process for both the \(x\) and \(y\) terms in the initial equation, you transform each into a squared form suitable for identifying the circle's center and radius.

Center of a Circle

In the context of a circle's equation, the center of the circle \((h, k)\) is the focal point from which all distances to the circumference are equal. Once you convert the circle's equation into its standard form through completing the square, you can easily identify these central coordinates.

For example, starting with:\[ (x - \frac{1}{4})^2 + (y + \frac{1}{4})^2 = \frac{3}{8} \]The parts of the equation \((x - \frac{1}{4})\) and \((y + \frac{1}{4})\) allow us to set the coordinates of the center of the circle. The expressions inside the parentheses directly suggest the values:

• The \(x\)-coordinate is \(\frac{1}{4}\).
• The \(y\)-coordinate is \(-\frac{1}{4}\), as the equation form becomes \((y - (-\frac{1}{4}))\).

The center \((\frac{1}{4}, -\frac{1}{4})\) serves as a reliable reference to plot the circle accurately on a graph.

Radius of a Circle

Understanding the radius of a circle is straightforward once the equation is in standard form. The radius is essentially the distance from the center to any point on the boundary of the circle, represented as \(r\) in the standard circle equation.

After completing the square, the equation:\[ (x - \frac{1}{4})^2 + (y + \frac{1}{4})^2 = \frac{3}{8} \]shows that

• \(r^2 = \frac{3}{8}\).
• Therefore, the radius \(r\) is simply \(\sqrt{\frac{3}{8}}\).

This calculated radius can help in many applications, including graphing or solving further geometric problems involving the circle. Always remember, the radius determines how big or small the circle is while defining its overall proportions.