Chapter 1: Problem 105
Show that the equation represents a circle, and find the center and radius of the circle. $$x^{2}+y^{2}-\frac{1}{2} x+\frac{1}{2} y=\frac{1}{8}$$
Short Answer
Expert verified
The circle has center \((\frac{1}{4}, -\frac{1}{4})\) and radius \(\sqrt{\frac{3}{8}}\).
Step by step solution
01
Rearrange the Equation
The given equation is \(x^{2} + y^{2} - \frac{1}{2} x + \frac{1}{2} y = \frac{1}{8}\). First, rearrange it to group the \(x\)- and \(y\)-terms on the left side: \(x^{2} - \frac{1}{2} x + y^{2} + \frac{1}{2} y = \frac{1}{8}\).
02
Complete the Square for x-terms
For the \(x\)-terms, complete the square. Take the coefficient of \(x\), which is \(-\frac{1}{2}\), divide it by 2, square the result: \((-\frac{1}{4})^{2} = \frac{1}{16}\). Add and subtract \(\frac{1}{16}\) inside the \(x\)-terms: \(x^{2} - \frac{1}{2}x + \frac{1}{16} - \frac{1}{16}\). The expression \((x^{2} - \frac{1}{2} x + \frac{1}{16})\) can now be written as \((x - \frac{1}{4})^{2}\).
03
Complete the Square for y-terms
For the \(y\)-terms, take the coefficient of \(y\), which is \(\frac{1}{2}\), divide by 2, and square it: \((\frac{1}{4})^{2} = \frac{1}{16}\). Add and subtract \(\frac{1}{16}\) inside the \(y\)-terms: \(y^{2} + \frac{1}{2}y + \frac{1}{16} - \frac{1}{16}\), which can be written as \((y + \frac{1}{4})^{2}\).
04
Simplify and Identify Circle Equation
Substitute back the completed squares into the equation: \((x - \frac{1}{4})^{2} - \frac{1}{16} + (y + \frac{1}{4})^{2} - \frac{1}{16} = \frac{1}{8}\). Add \(\frac{1}{16}\) twice to both sides to balance the equation: \((x - \frac{1}{4})^{2} + (y + \frac{1}{4})^{2} = \frac{1}{8} + \frac{1}{16} + \frac{1}{16} = \frac{2}{8} + \frac{1}{8} = \frac{4}{16} + \frac{2}{16} = \frac{6}{16} = \frac{3}{8}\).
05
Identify Center and Radius
The equation \((x - \frac{1}{4})^{2} + (y + \frac{1}{4})^{2} = \frac{3}{8}\) represents a circle. The center of the circle is \((\frac{1}{4}, -\frac{1}{4})\) and the radius is the square root of \(\frac{3}{8}\), which is \(\sqrt{\frac{3}{8}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the square
Completing the square is an essential method used to transform a quadratic equation into a perfect square trinomial. This technique is particularly helpful when working with equations of a circle, as it allows us to rewrite them in a form that easily reveals their geometric properties. Let's take a look at how to apply this method.
- First, focus on the quadratic terms, such as the terms involving only x or y.
- Extract the coefficient of the linear term (the term with just x or y), divide it by two, and then square the result.
- Introduce this squared term to both sides of the equation, ensuring that the balance is maintained.
Center and radius of a circle
The center and radius of a circle provide vital information about its position and size in coordinate geometry. The equation of a circle in standard form is expressed as:\((x - h)^2 + (y - k)^2 = r^2\)where:
h andk represent the coordinates of the centerr is the radius of the circle
Equation of a circle
The equation of a circle is typically written in standard form as \((x - h)^2 + (y - k)^2 = r^2\),representing all points \((x, y)\)that are a distance r from the center \((h, k)\).
Steps to identifying and using this form:
- Whenever given an equation that looks quadratic in both x and y, consider if it represents a circle.
- Rearrange and complete the square for both variables to easily transition it into the standard form.
- Compare your equation with the standard circle equation format to extract specific details like the center and the radius.