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Magazine Chapter 2: Problem 72
65鈥�72 ? Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}+\frac{1}{2} x+2 y+\frac{1}{16}=0 $$
Short Answer
Expert verified
The circle has center (-1/4, -1) and radius 1.
Step by step solution
01
Identify the Circle Equation Template
A circle in standard form is expressed as \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
02
Group and Rearrange Terms
Rearrange and group the terms of \(x\) and \(y\) in the equation: \[x^2 + \frac{1}{2}x + y^2 + 2y = -\frac{1}{16}.\]
03
Complete the Square for x
To complete the square for \(x\), take the coefficient of \(x\), \(\frac{1}{2}\), halve it to get \(\frac{1}{4}\), square it to get \(\frac{1}{16}\), then add and subtract \(\frac{1}{16}\) to balance the equation: \[x^2 + \frac{1}{2}x = \left(x + \frac{1}{4}\right)^2 - \frac{1}{16}.\]
04
Complete the Square for y
Perform a similar operation for \(y\): take the coefficient of \(y\), \(2\), halve it to get \(1\), square it to get \(1\), then add and subtract \(1\) to balance the equation: \[y^2 + 2y = (y + 1)^2 - 1.\]
05
Combine and Simplify
Substitute the completed square forms back into the equation: \[(x + \frac{1}{4})^2 - \frac{1}{16} + (y+1)^2 - 1 = -\frac{1}{16}.\]Combine and balance both sides of the equation:\[(x + \frac{1}{4})^2 + (y+1)^2 = \frac{1}{16} + 1 - \frac{1}{16}.\]This simplifies to:\[(x + \frac{1}{4})^2 + (y+1)^2 = 1.\]
06
Identify the Center and Radius
Now that the equation is in standard form \((x-h)^2 + (y-k)^2 = r^2\), identify the center, \((-\frac{1}{4}, -1)\), and radius, \(1\), since the equation simplifies to \(r^2=1\).
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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 a technique often used in algebra. It reorganizes a quadratic equation into a perfect square trinomial, making it easier to solve or transform. In the context of circle equations, this method helps rewrite the equation of a circle in standard form.
For a single term like \(x^2 + \frac{1}{2}x\), you must take:
For a single term like \(x^2 + \frac{1}{2}x\), you must take:
- Half of the linear coefficient (\(\frac{1}{2}\) divided by 2 equals \(\frac{1}{4}\))
- Square it (\((\frac{1}{4})^2 = \frac{1}{16}\))
- Add and subtract this square (\(\frac{1}{16}\)) to complete the square
Standard Form of a Circle
The "standard form of a circle" offers a neat way to represent a circle's equation. This form is shown as \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h, k)\) determines the center of the circle, and \(r\) is the radius.
This structure allows for easy identification of the circle's characteristics:
This structure allows for easy identification of the circle's characteristics:
- It sets both the x and y terms into perfect squares, reflecting the circle's symmetrical nature
- Provides a straightforward method to read off the circle's coordinates and size
- Transforms complex equations into simple central representations
Circle Center
The center of a circle, denoted typically by \((h, k)\), is the point from which every point on the circle is equidistant. In the standard circle equation form \((x-h)^2 + (y-k)^2 = r^2\), the values for \(h\) and \(k\) are derived directly from the completed square terms.
In our example, after completing the square, the equation becomes \((x + \frac{1}{4})^2 + (y + 1)^2 = 1\). From this, we can observe:
In our example, after completing the square, the equation becomes \((x + \frac{1}{4})^2 + (y + 1)^2 = 1\). From this, we can observe:
- The horizontal shift from 0 is \(-\frac{1}{4}\), making \(h = -\frac{1}{4}\)
- The vertical shift from 0 is \(-1\), making \(k = -1\)
Circle Radius
The radius of a circle is a measure of distance from the center to any point on the circle's edge. Upon transforming the circle's equation into its standard form \((x-h)^2 + (y-k)^2 = r^2\), finding the radius becomes straightforward.
From our example, the equation is simplified to \((x + \frac{1}{4})^2 + (y + 1)^2 = 1\). The right side of the equation equals \(r^2\), meaning the radius squared:
From our example, the equation is simplified to \((x + \frac{1}{4})^2 + (y + 1)^2 = 1\). The right side of the equation equals \(r^2\), meaning the radius squared:
- Since \(r^2 = 1\), the radius \(r = \sqrt{1}\)
- Therefore, \(r = 1\)
