Problem 64 - Precalculus 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 standard form of the given equation is \[(x + \frac{3}{2})^{2} + (y + \frac{5}{2})^{2} = \frac{25}{4}\]. The center of the circle is \((- \frac{3}{2}, - \frac{5}{2})\) and the radius is \(\frac{5}{2}\).

Step 1: Group The x and y Coefficients Separately

Rearrange the terms of \(x\) and \(y\) separately: \[(x^2 + 3x) + (y^2 + 5y) = -\frac{9}{4}\].

Step 2: Completing the Square for x and y Terms

Complete the square on the \(x\) terms and the \(y\) terms. Half of the coefficient of \(x\) is \(\frac{3}{2}\), and \((\frac{3}{2})^2 = \frac{9}{4}\). Half of the coefficient of \(y\) is \(\frac{5}{2}\), and \((\frac{5}{2})^2 = \frac{25}{4}\). So the equation becomes:\[(x^2 + 3x + \frac{9}{4}) + (y^2 + 5y + \frac{25}{4}) = -\frac{9}{4} + \frac{9}{4} + \frac{25}{4}\].

Step 3: Simplify Form to Standard Form

Simplify the equation by writing each bracket as a square and simplifying the right-hand side: \[(x + \frac{3}{2})^2 + (y + \frac{5}{2})^2 = \frac{25}{4}\]. This is the standard form of a circle.

Step 4: Find the center (h, k) and radius r of the circle

The center is determined by \((- \frac{3}{2}, - \frac{5}{2})\) and the radius is the square root of the right side, i.e., \(r = \frac{5}{2}\).

Step 5: Graph the Equation

Plot the center point at \((- \frac{3}{2}, - \frac{5}{2})\). Draw a circle with radius \(r = \frac{5}{2}\) centered on this point.

Key Concepts

Standard Form of a Circle

The standard form of a circle is a way of expressing the equation of a circle that makes it easy to identify the circle's center and radius. It is especially helpful when you want to analyze or graph a circle. In general, the standard form equation of a circle is written as:

• \((x - h)^2 + (y - k)^2 = r^2\)

Here,

• \( (x - h)^2 \) and \( (y - k)^2 \) represent the squared distances from every point on the circle to the center \((h, k)\).
• \( r\) is the radius of the circle.

To convert any general equation of a circle into this standard form, you'll most often need to "complete the square" for both the \(x\) and \(y\) terms. This process allows you to express the squared terms neatly, facilitating the identification of the circle's geometry.

Circle Equation

Any circle can be represented by a specific equation which depends on its radius and the coordinates of its center. Circles are unique as they have constant distance from a fixed point, the center. In mathematics, this is precisely defined by its equation.
In cases where the circle’s equation is provided, like in the given exercise, starting from \[ x^2 + y^2 + 3x + 5y + \frac{9}{4} = 0 \],
restructuring it in standard form involves:

• Group the \(x\) and \(y\) terms together separately.
• Use completing the square to form perfect square trinomials.
• Adjust the equation so it isolates the squared terms on one side of the equation in the form \((x + a)^2 + (y + b)^2 = c^2\).

This method transforms the complicated equation into something recognizable and manageable, making it clear how the circle is shaped and where it is positioned within a coordinate system.

Center and Radius of a Circle

The center and radius are two vital attributes of a circle, and can be easily determined from the circle's equation when it is in standard form. Using the form \((x-h)^2 + (y-k)^2 = r^2\), you can instantly identify:

• Center (h, k): The values of \(h\) and \(k\) are the coordinates of the center of the circle. They are the constants subtracted from the \(x\) and \(y\) terms respectively.
• Radius r: The radius is given by \(r\), which is the square root of the value on the right side of the equation. It represents the distance from the center of the circle to any point on the circumference.

For the exercise at hand, converting the circle equation \((x + \frac{3}{2})^2 + (y + \frac{5}{2})^2 = \frac{25}{4}\), reveals:

• The center is at \((- \frac{3}{2}, - \frac{5}{2})\).
• The radius is \(\frac{5}{2}\), because the right-hand value \(\frac{25}{4}\) when square-rooted provides the radius.

Understanding how to quickly decipher these attributes from the equation allows for easier plotting and understanding of a circle's position and size in a graph.