Problem 27 - Algebra for College Students

Question

For Problems \(15-32\), find the center and the length of a radius of each of the circles. $$ x^{2}+y^{2}-5 y-1=0 \quad\left(0, \frac{5}{2}\right): r=\frac{\sqrt{29}}{2} $$

Step-by-Step Solution

Answer

Center: \((0, \frac{5}{2})\); Radius: \(\frac{\sqrt{29}}{2}\).

Step 1: Rewrite the Circle Equation

The given equation of the circle is \( x^2 + y^2 - 5y - 1 = 0 \). We can rewrite this equation in the standard form of a circle equation \( x^2 + (y - k)^2 = r^2 \) by completing the square for the \( y \)-terms.

Step 2: Complete the Square for y

To complete the square for \( y \), isolate the \( y \)-terms: \( y^2 - 5y \). Complete the square by adding and subtracting \( \left(\frac{5}{2}\right)^2 \), which equals \( \frac{25}{4} \). Hence, we rewrite \( y^2 - 5y \) as \( (y - \frac{5}{2})^2 - \frac{25}{4} \).

Step 3: Adjust Equation with Completed Square

Substitute the completed square form back into the original equation: \( x^2 + (y - \frac{5}{2})^2 - \frac{25}{4} - 1 = 0 \). Simplify to \( x^2 + (y - \frac{5}{2})^2 = 1 + \frac{25}{4} \).

Step 4: Simplify and Find Radius

Combine the constants: \( 1 + \frac{25}{4} = \frac{4}{4} + \frac{25}{4} = \frac{29}{4} \). This gives us \( x^2 + (y - \frac{5}{2})^2 = \frac{29}{4} \), which represents the standard form \( (x - 0)^2 + (y - \frac{5}{2})^2 = \left(\frac{\sqrt{29}}{2}\right)^2 \). The radius \( r \) is \( \frac{\sqrt{29}}{2} \).

Step 5: Identify the Center and Radius

From the equation \( (x - 0)^2 + (y - \frac{5}{2})^2 = \left(\frac{\sqrt{29}}{2}\right)^2 \), it is clear that the center of the circle is \((0, \frac{5}{2})\), and the radius is \( \frac{\sqrt{29}}{2} \).

Key Concepts

Completing the Square

To convert a quadratic equation into a form that reveals the features of a circle, we often use a method called "Completing the Square." This technique helps us turn a standard quadratic expression into a perfect square trinomial. Let's see how it works for the \(y\) terms.

In the example \( x^2 + y^2 - 5y - 1 = 0 \), we want to focus on the \(y\) part, especially \( y^2 - 5y \). Completing the square means converting this part to the form \((y - a)^2\), which is much easier to manage.

1. Start by isolating \(y\) terms: \( y^2 - 5y \).

• Take the coefficient of \(y\), which is \(-5\), and halve it, giving us \(-\frac{5}{2}\).
• Square this value to get \(\left(\frac{5}{2}\right)^2 = \frac{25}{4}\).

2. Add and subtract \(\frac{25}{4}\) within the equation to balance it out.

• This changes \(y^2 - 5y\) to \((y - \frac{5}{2})^2 - \frac{25}{4}\).

Now, by completing the square, we mold the expression into a format where it can be neatly incorporated back into solving the equation.

Standard Form of a Circle

The standard form of a circle's equation makes it easy to identify both the center and the radius of the circle. It is generally written as:\[(x - h)^2 + (y - k)^2 = r^2\] where \((h, k)\) is the center of the circle, and \( r \) is its radius.

By using the completed square from the previous step, the original circle equation \( x^2 + y^2 - 5y - 1 = 0 \) is rewritten. Substituting the results we obtained gives:\[x^2 + (y - \frac{5}{2})^2 = \frac{29}{4}\]This corresponds to the format \((x - 0)^2 + (y - \frac{5}{2})^2 = \left(\frac{\sqrt{29}}{2}\right)^2\).

Notice:

• The term \((x - 0)^2\) indicates that \( h = 0 \).
• \((y - \frac{5}{2})^2\) indicates that \( k = \frac{5}{2} \).
• Finally, the constant \( \left(\frac{\sqrt{29}}{2}\right)^2 = \frac{29}{4}\) reveals the square of the radius.

Oddly enough, deploying a standard form makes life easier for solving and visualizing the geometric properties of the circle.

Center and Radius of Circle

Identifying the center and radius of a circle becomes quite straightforward once the circle's equation is written in standard form. The center \((h, k)\) pinpoints the exact spot around which the circle is perfectly symmetrical.

From the equation \((x - 0)^2 + (y - \frac{5}{2})^2 = \left(\frac{\sqrt{29}}{2}\right)^2\), we see:

• The center is at \((0, \frac{5}{2})\).
• The radius—is the distance from this center to any point on the circle—is \( \frac{\sqrt{29}}{2} \).

The information provided allows us to not only sketch the circle on a coordinate plane but also to understand how this circle is positioned without graphing. It tells us that:

• The circle is perfectly horizontal and vertical in symmetry about the center \((0, \frac{5}{2})\).
• The radius, extending \( \frac{\sqrt{29}}{2} \) units from the center, defines its size.

By knowing both the center and radius, we can draw or imagine the circle accurately in space. This understanding is fundamental in both algebra and geometry, leading to a deeper grasp of how equations translate into geometric figures.