Problem 72 - Algebra and Trigonometry

Question

65–72 ? Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}+\frac{1}{2} x+2 y+\frac{1}{16}=0 $$

Step-by-Step Solution

Answer

The circle has center (-1/4, -1) and radius 1.

Step 1: 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.

Step 2: Group and Rearrange Terms

Rearrange and group the terms of \(x\) and \(y\) in the equation: \[x^2 + \frac{1}{2}x + y^2 + 2y = -\frac{1}{16}.\]

Step 3: 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: \[x^2 + \frac{1}{2}x = \left(x + \frac{1}{4}\right)^2 - \frac{1}{16}.\]

Step 4: 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: \[y^2 + 2y = (y + 1)^2 - 1.\]

Step 5: 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:\[(x + \frac{1}{4})^2 + (y+1)^2 = 1.\]

Step 6: 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 \(r^2=1\).

Key Concepts

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:

• 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

Now the term \(x^2 + \frac{1}{2}x\) can be expressed as \((x + \frac{1}{4})^2 - \frac{1}{16}\). This method simplifies the equation and sets it up nicely for finding the circle's center and radius.

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:

• 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

Any complex quadratic equation representing a circle can be converted to this form using methods like "completing the square," making it immensely useful in geometric analysis.

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:

• The horizontal shift from 0 is \(-\frac{1}{4}\), making \(h = -\frac{1}{4}\)
• The vertical shift from 0 is \(-1\), making \(k = -1\)

Thus, the center of the circle can be simply read as \((-\frac{1}{4}, -1)\). This assures that the equation is precisely oriented around this central point.

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:

• Since \(r^2 = 1\), the radius \(r = \sqrt{1}\)
• Therefore, \(r = 1\)

The radius tells you not only about the size of the circle but also confirms its symmetry. Each point along the perimeter is exactly 1 unit away from the center \((-\frac{1}{4}, -1)\), encapsulating the balance inherent in a circle's form.