Problem 54 - A Graphical Approach to College Algebra

Question

Decide whether each equation has a circle as its graph. If it does, give the center and radius. $$4 x^{2}+4 x+4 y^{2}-4 y-3=0$$

Step-by-Step Solution

Answer

It's a circle, centered at \((-\frac{1}{2}, \frac{1}{2})\) with radius \(\sqrt{\frac{5}{4}}\).

Step 1: Arrange the Equation

First, we notice that the given equation is not yet in the standard form of a circle equation. A circle's equation is typically in the form: \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Let's start by arranging terms: \[4x^2 + 4x + 4y^2 - 4y = 3\]

Step 2: Factor Out Constants

Next, factor out the common factor of 4 from the terms involving \(x\) and \(y\):\[4(x^2 + x + y^2 - y) = 3\]

Step 3: Complete the Square for x and y

Now we complete the square for both \(x\) and \(y\):For the \(x\) terms, in \(x^2 + x\), add and subtract \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\):\[x^2 + x = (x + \frac{1}{2})^2 - \frac{1}{4}\]For the \(y\) terms in \(y^2 - y\), add and subtract \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\):\[y^2 - y = (y - \frac{1}{2})^2 - \frac{1}{4}\]

Step 4: Substitute Back into the Equation

Substitute the completed square terms back into the equation:\[4((x + \frac{1}{2})^2 - \frac{1}{4} + (y - \frac{1}{2})^2 - \frac{1}{4}) = 3\]

Step 5: Simplify the Equation

Distribute the 4 and move constants to the right side:\[4((x + \frac{1}{2})^2 + (y - \frac{1}{2})^2) = 3 + 1 + 1\]\[4((x + \frac{1}{2})^2 + (y - \frac{1}{2})^2) = 5\]

Step 6: Divide to Standard Circle Form

Finally, divide the entire equation by 4 to match the standard form of a circle:\[(x + \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{5}{4}\]Thus, the equation is now in the form of a circle.

Key Concepts

Completing the Square

Completing the square is a method used to transform quadratic equations into a perfect square trinomial. It is especially useful when trying to convert equations into a form that is easier to graph or interpret. In the context of circles, completing the square helps us rewrite general quadratic equations in a way that reveals the center and radius of the circle.

To complete the square, you start by focusing on the quadratic and linear terms of either the variable, typically in the form of expressions like:

• \(x^2 + bx\) or \(y^2 - by\).

Here’s a guided process:

• Take the coefficient of the linear term, divide it by two, and then square it.
• Add and subtract this squared value within the equation. This is essential as it does not change the equation’s overall balance.
• Group the terms so they form a perfect square trinomial.
• Write the trinomial as a binomial squared.

An example from our problem: We manipulate \(x^2 + x\) by adding and subtracting \(\left(\frac{1}{2}\right)^2\) to get:

• \((x + \frac{1}{2})^2 - \frac{1}{4}\).

This new expression represents a perfect square, which is crucial for identifying the structure of a circle's equation.

Standard Form of a Circle

The standard form of a circle's equation provides a straightforward way to determine a circle's key characteristics, such as its center and radius. This form is expressed as:

• \((x-h)^2 + (y-k)^2 = r^2\).

In this equation:

• \((h, k)\) represents the circle's center, making it easy to identify the location of the circle on a coordinate plane.
• \(r\) stands for the radius, which tells how far points on the circle are from the center.

Turning a general equation into this standard form involves rearranging terms and using techniques like completing the square.

• This standard form makes it easy to visually identify circles on a graph.
• In comparison to a more complex equation, the standard form clearly presents the circle’s essential features.

For instance, in the processed equation \((x + \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{5}{4}\), we can directly see the center as \((-\frac{1}{2}, \frac{1}{2})\) and the radius as \(\sqrt{\frac{5}{4}}\).

Center and Radius of a Circle

Identifying the center and radius of a circle from its equation is essential for understanding its geometry. Once an equation is transformed into the standard form \((x-h)^2 + (y-k)^2 = r^2\), extracting this information becomes straightforward.

Here is how you can easily identify these components:

• The center of the circle is directly taken from the values of \(h\) and \(k\) in the equation.
• These values correspond to \(x+h=0\) or \(x-h=0\) shift for the horizontal axis and \(y+k=0\) or \(y-k=0\) shift for the vertical axis.
• To find the radius \(r\), take the square root of the number on the right of the equation.

For example, the equation \((x + \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{5}{4}\) tells us:

• The center of the circle is at \((-\frac{1}{2}, \frac{1}{2})\), noticing the signs have flipped from the equation terms \((x+h)\) and \((y-k)\).
• The radius is \(\sqrt{\frac{5}{4}}\), which simplifies to \(\frac{\sqrt{5}}{2}\).

Understanding these components allows you to visualize the circle’s position and size on the coordinate plane effectively.