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 \((-\frac{1}{4}, -1)\) and radius 1.

Step 1: Identify the General Form of a Circle

The general form of a circle is \( x^2 + y^2 + ax + by + c = 0 \). The given equation \( x^2 + y^2 + \frac{1}{2} x + 2y + \frac{1}{16} = 0 \) is in this form, suggesting it represents a circle.

Step 2: Rearrange Terms for Completion of Square

To find the center and radius, we need to complete the square. Group the \( x \) and \( y \) terms together: \[ x^2 + \frac{1}{2}x + y^2 + 2y = -\frac{1}{16} \]

Step 3: Complete the Square for Terms Involving x

Take the coefficient of \( x \) (which is \( \frac{1}{2} \)), halve it, and square it: \[ \left( \frac{1}{4} \right)^2 = \frac{1}{16} \]Add and subtract \( \frac{1}{16} \) in the equation: \[ x^2 + \frac{1}{2}x + \frac{1}{16} - \frac{1}{16} + y^2 + 2y = -\frac{1}{16} \]

Step 4: Complete the Square for Terms Involving y

Take the coefficient of \( y \) (which is \( 2 \)), halve it, and square it: \[ 1^2 = 1 \]Add and subtract \( 1 \) in the equation:\[ x^2 + \frac{1}{2}x + \frac{1}{16} + y^2 + 2y + 1 - 1 = -\frac{1}{16} + \frac{1}{16} \]

Step 5: Rewrite the Equation in Completed Square Form

The equation can be rewritten as:\[ \left(x + \frac{1}{4}\right)^2 + (y + 1)^2 = 1 \]This is now in the standard form of a circle \( (x-h)^2 + (y-k)^2 = r^2 \).

Step 6: Identify Center and Radius

From the standard form \( \left(x + \frac{1}{4}\right)^2 + (y + 1)^2 = 1 \), we identify the center as \( \left(-\frac{1}{4}, -1\right) \) and the radius as \( 1 \).

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This method is particularly useful when dealing with equations of conic sections, like circles. The goal is to write the quadratic terms as a square of a binomial, making it easier to identify key properties of the equation, such as the center and radius of a circle.

• First, focus on one variable at a time, typically beginning with the terms involving 'x'.
• To complete the square for the 'x' terms, take the coefficient of 'x', halve it, and then square it.
• Add and subtract this squared value within the equation to keep the balance.
• Repeat this process for the 'y' terms in the equation.

For instance, in the given equation, the 'x' terms: \[ x^2 + \frac{1}{2}x \] require the completion. Take \( \frac{1}{2} \), halve it to \( \frac{1}{4} \), and square it to get \( \frac{1}{16} \). By adding and subtracting \( \frac{1}{16} \), we ensure the equation remains equivalent, allowing us to rewrite as a squared term: \[ \left(x + \frac{1}{4}\right)^2 \].
This step is mirrored for the 'y' terms to achieve a similar transformed expression. Completing the square is essential as it helps to clearly identify the circle's defining characteristics.

Center of a Circle

The center of a circle in a two-dimensional plane is a fixed point that is equidistant from any point along the circumference of the circle. Finding the center from a given circle equation involves a clear understanding of its standard form: \[(x-h)^2 + (y-k)^2 = r^2\]Here, \((h, k)\) represents the coordinates of the circle's center. We obtain these values through the process of completing the square.

• Identify the form \((x - h), (y - k)\) within the rewritten equation.
• The values of \(h\) and \(k\) determine the exact location of the center.

In our example, after rewriting the equation as a perfect square:\[ \left(x + \frac{1}{4}\right)^2 + (y + 1)^2 = 1\] We identify \(h\) as \(-\frac{1}{4}\) and \(k\) as \(-1\), giving the center \((-\frac{1}{4}, -1)\).
Understanding the center is essential in analyzing a circle's geometry and solving related problems like graphing or determining intersections.

Radius of a Circle

The radius of a circle is the distance from the circle's center to any point on its edge. It is a key component of the circle's geometry and is represented by \(r\) in the standard circle equation.To find the radius from an equation, we look at the format:\[(x-h)^2 + (y-k)^2 = r^2\]Here, \(r^2\) clearly shows the square of the radius.

• Take the equation in its final perfect square form.
• Identify the value set equal to the binomials, which represents \(r^2\).
• The radius \(r\) is then the positive square root of this value.

In our completed square equation, \[ \left(x + \frac{1}{4}\right)^2 + (y + 1)^2 = 1\] we see that \(r^2\) is equal to \(1\). Taking the square root, we find \(r = 1\).
Thus, the radius signifies how wide the circle is. This information, along with the center, can be used for graphing and to describe the circle in various applications.