Problem 64 - College Algebra Essentials

Question

complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}+y^{2}+3 x+5 y+\frac{9}{4}=0 $$

Step-by-Step Solution

Answer

The center of the circle is (-1.5, -2.5) and the radius is 2.5.

Step 1: Structural Adjustment

Rearrange the given equation and group the x and y terms separately: \( x^{2}+3x + y^{2}+5y = -\frac{9}{4} \)

Step 2: Complete the Square

To complete squares for x and y parts separately, half the coefficient of x and y (which are 3 and 5 respectively), square them and add to both sides of the equation. This means adding \((\frac{3}{2})^{2} = \frac{9}{4}\) to the x part and adding \((\frac{5}{2})^{2} = \frac{25}{4}\) to the y part, leading to: \( x^{2}+3x + \left(\frac{3}{2}\right)^{2} + y^{2}+5y + \left(\frac{5}{2}\right)^{2} = -\frac{9}{4} + \frac{9}{4} + \frac{25}{4} \)

Step 3: Simplify and Rewrite in Standard Form

After adding, simplify to obtain: \( (x+\frac{3}{2})^{2} + (y+\frac{5}{2})^{2} = \frac{25}{4} \). This is the equation in standard form.

Step 4: Identify Center and Radius

The center of the circle is given by the terms -(h,k) in the standard form. So, here the center is \(-(\frac{3}{2}, \frac{5}{2}) = (-1.5, -2.5)\). The radius is the square root of the right hand side of the equation, so \( r=\frac{5}{2} = 2.5 \)

Step 5: Graph the Circle

Now, graph the equation by drawing a circle centered at (-1.5, -2.5), with a radius of 2.5.

Key Concepts

Completing the Square

Completing the square is a technique used in algebra to transform a quadratic equation into a perfect square trinomial. This method helps in rewriting equations in a form that is easily interpretable, for solving or graphing.
To complete the square, focus on the quadratic terms separately for each variable. For example, with terms like \( x^2 + 3x \), you want to reshape it into the form \( (x + A)^2 \).

• First, take half of the coefficient of \(x\) or \(y\), and square it. Here, the coefficient for \(x\) is 3, so half of it is \(\frac{3}{2}\) and squaring it gives \(\frac{9}{4}\).
• Do the same for the \(y\) term with 5, resulting in \((\frac{5}{2})^2 = \frac{25}{4}\).
• Add these squared terms inside the equation to make it a complete square trinomial.

This method helps isolate the circle's properties, making it easier to interpret the equation geometrically.

Standard Form of a Circle

The standard form of a circle's equation reveals its most important features, making it much easier to graph and understand the circle's geometry right away. The standard form is written as:
\[(x-h)^2 + (y-k)^2 = r^2\]
Here, \( (h, k) \) represents the center of the circle, and \( r \) is the radius. This form highlights these features clearly, providing almost instant insight into the circle's position and size.
In the exercise, after completing the square, we turned the original equation into its standard form:
\[(x + \frac{3}{2})^2 + (y + \frac{5}{2})^2 = \left(\frac{5}{2}\right)^2\]
This rewriting process not only helps in visual interpretation but also simplifies solving related algebraic problems.

Center of a Circle

The center of a circle is a vital component of its geometric shape. In the standard form \((x-h)^2 + (y-k)^2 = r^2\), the center is located at \((h, k)\). Knowing the center helps plot the circle accurately on a coordinate plane.
In our transformed equation, \((x + \frac{3}{2})^2 + (y + \frac{5}{2})^2 = \frac{25}{4}\), the center will be the opposite of what lies next to \(x\) and \(y\):

• The center \( (h, k) \) is \((-\frac{3}{2}, -\frac{5}{2})\), which simplifies to \((-1.5, -2.5)\).

This position is crucial for graphing the circle, ensuring it is drawn accurately in its exact spot on a graph.

Radius of a Circle

The radius is the distance from the center of the circle to any point on its circumference. In the equation \((x-h)^2 + (y-k)^2 = r^2\), \(r^2\) refers to the square of the radius. Thus, taking the square root of both sides of the equation provides the actual radius value.
In this exercise, the right-hand side of the standard form equation is \(\frac{25}{4}\):

• The radius is \(\sqrt{\frac{25}{4}} = \frac{5}{2}\).
• This measures 2.5 units, designating how far the circle extends from its center \((-1.5, -2.5)\).

Understanding the radius is essential for drawing the circle correctly, ensuring that its size is neither too small nor too large when graphed on a coordinate plane.