Problem 64 - College Algebra

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: Rearrange the equation

Rearrange the terms: group x's and y's together and move constant numbers to the other side of the equation: \[x^{2}+3x+y^{2}+5y = -\frac{9}{4}\]

Step 2: Apply completing the square

Complete the square for the x's and y's. Half of 3 is \( \frac{3}{2}\), and squaring it gives \( \frac{9}{4}\), similarly half of 5 is \( \frac{5}{2} \) and squaring it gives \( \frac{25}{4}\). Add these squares to both sides of equation: \[(x^{2} + 3x + \frac{9}{4}) + (y^{2} + 5y + \frac{25}{4}) = -\frac{9}{4} + \frac{9}{4} + \frac{25}{4} \]

Step 3: Simplify the equation to its standard form

Simplify the equation to its standard form: \[((x + \frac{3}{2})^{2} + (y + \frac{5}{2})^{2}) = 6.25\]

Step 4: Identify circle's center and radius

This is now in the standard form for a circle's equation, where the center of the circle is at (-a, -b) and the radius r is the square root of the right side of equation. So, the center of circle is \((- \frac{3}{2}, - \frac{5}{2})\) and radius is \( \sqrt{6.25}= 2.5\)

Key Concepts

Equation of a Circle

The equation of a circle is a fascinating mathematical expression that can be derived and manipulated through the process of completing the square, as shown in the original exercise. Generally, the equation of a circle is defined as \[(x - h)^2 + (y - k)^2 = r^2\] where

• \(h\) and \(k\) are the coordinates of the center of the circle, \((h, k)\),
• \(r\) is the radius of the circle.

Completing the square is essential when the circle’s equation does not initially appear in the standard form. In the exercise, completing the square allows the equation to shift from \(x^2 + y^2 + 3x + 5y + \frac{9}{4} = 0\) to \((x + \frac{3}{2})^2 + (y + \frac{5}{2})^2 = 6.25\), clearly identifying the circle's center \((-\frac{3}{2}, -\frac{5}{2})\) and its radius \(2.5\).
Understanding this process helps in identifying the geometrical properties of a circle from a mere algebraic expression.

Standard Form

The standard form of a circle's equation provides a straightforward way to identify its key features: the center and radius. Typically written as \((x - h)^2 + (y - k)^2 = r^2\), this formula directly reveals the circle's center \((h, k)\) and radius \(r\).
In the given exercise, transforming the equation into standard form involved a few crucial steps.

• Firstly, terms involving \(x\) and \(y\) were isolated and rearranged.
• Then, completing the square was applied to both the \(x\) and \(y\) terms, yielding \(\frac{9}{4}\) and \(\frac{25}{4}\), respectively.
• Finally, the equation was simplified to \((x+\frac{3}{2})^2+(y+\frac{5}{2})^2 = 6.25\).

This standard form not only makes it easy to graph but also facilitates understanding geometric relationships and transformations involving circles. It plays a vital role in analytical geometry and helps students visualize the position and size of a circle based purely on algebraic transformations.

Graphing Circles

Once the equation of a circle is in standard form, graphing it becomes quite intuitive and straightforward. The center of the circle can easily be plotted using the coordinates \(h\) and \(k\) derived from the equation \((x - h)^2 + (y - k)^2 = r^2\). In the exercise, by identifying \((-\frac{3}{2}, -\frac{5}{2})\) as the center and \(2.5\) as the radius, we can plot the circle accurately on a graph.

• Start by marking the center point at \((-\frac{3}{2}, -\frac{5}{2})\) on the Cartesian plane.
• Then, using a compass or an estimated measurement based on the scale, draw the circle with a radius of \(2.5\), equally extending from the center in all directions.

Graphing the circle helps visualize its size and location relative to the coordinate axes, providing a clearer understanding of its geometrical properties. This visual representation is beneficial in numerous applications, from design to problem-solving in mathematical contexts.