Problem 98 - College Algebra and Calculus: An Applied Approach

Question

Write the equation of the circle in standard form. Then sketch the circle. \(4 x^{2}+4 y^{2}-4 x+2 y-1=0\)

Step-by-Step Solution

Answer

The standard form of the circle's equation is \((x-0.5)^{2}+(y+0.25)^{2}=1\), the center is at point (0.5,-0.25), and the radius is 1.

Step 1: Re-arrange the equation into standard form

Firstly divide the given equation by 4 to simplify it. The equation becomes: \(x^{2}-x+y^{2}+\frac{y}{2}-\frac{1}{4}=0\), which should be rearranged into the form \(x^{2}-2hx+y^{2}-2ky=-1\). Combining like terms and completing the square, we get \((x-0.5)^{2}+(y+0.25)^{2}=1\). So according to this representation, h=0.5, k=-0.25, and r=1.

Step 2: Identify the circle's center and radius

From the standard form equation we can directly read off that the center is at point (h,k)=(0.5,-0.25) and the radius r is 1.

Step 3: Sketch the circle

Using the values from step 2, place the point (0.5,-0.25) at the center and then measure 1 unit in each direction. Draw the circle from these points.

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic expression into a perfect square trinomial by adding and subtracting a certain value. When dealing with circles in algebra, this technique is particularly useful for writing the equation of a circle in standard form.

Consider a quadratic equation in the format of \(ax^2 + bx + c\). To complete the square, the steps involve:

• Dividing the entire equation by 'a' if 'a' is not equal to 1, to make the coefficient of \(x^2\) term 1.
• Finding a value that will create a perfect square trinomial from the \(x\) terms, which is \(\frac{b}{2a}\)^2
• Adding and subtracting this value within the equation to maintain equality.
• Factor the perfect square trinomial and simplify the equation as needed.

In the context of our problem, we divided the equation by 4 and then completed the square for both the \(x\) and \(y\) terms, thereby rearranging the equation into a recognizable circle equation.

Circle Graph Sketching

Once the standard form of the circle equation is obtained through completing the square, sketching the graph becomes an easier task. The standard form for a circle's equation is \( (x-h)^{2} + (y-k)^{2} = r^2 \) where \( (h,k) \) is the center of the circle and \(r\) is its radius.

To sketch the graph, follow these steps:

• Locate the center \( (h,k) \) on the coordinate plane.
• From the center, move a distance equal to the radius along the horizontal and vertical directions to identify points on the circle.
• Using these points as a guide, sketch the circle as a round curve that is equidistant from the center at every point.

In our example, after finding the standard form, we identified the center at \( (0.5,-0.25) \) with a radius of 1, which facilitated the sketching process.

Circles in Algebra

Circles are geometric shapes that can be represented algebraically with equations. Understanding the algebra of circles is crucial for solving problems involving circle graphs and their properties. In algebra, a common representation of a circle is the standard form equation \( (x-h)^{2} + (y-k)^{2} = r^2 \), where \( (h,k) \) denotes the circle's center and \(r\) represents the radius.

The algebra links directly to geometric interpretations, allowing us to infer the size and position of the circle just by examining its equation. In algebraic operations, changes to these parameters transform the circle in various ways—shifting, resizing, and even reflecting it across axes.

Standard Form Conversion

Standard form conversion involves rearranging an equation into a widely recognized and usable format. The standard form for the equation of a circle, \( (x-h)^{2} + (y-k)^{2} = r^2 \) offers clarity in identifying the center and radius of the circle, crucial for graphing and solving problems related to a circle's geometry. Converting to standard form typically involves completing the square and making algebraic manipulations so that the equation reflects the necessary parameters of ‘h’, ‘k’, and ‘r’.

Our example underwent the conversion process, beginning from a general quadratic expression and restructuring it into the standard form by completing the square. This conversion aims for simplicity and immediacy in using the equation to derive geometric properties of the circle.