Problem 30 - Algebra for College Students

Question

For Problems \(15-32\), find the center and the length of a radius of each of the circles. $$ 4 x^{2}+4 y^{2}=1 \quad(0,0), r=\frac{1}{2} $$

Step-by-Step Solution

Answer

Center: (0,0); Radius: \(\frac{1}{2}\).

Step 1: Identify the Standard Equation of a Circle

The standard form of the equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius.

Step 2: Simplify the Given Equation

The given equation is \(4x^2 + 4y^2 = 1\). We need to divide the entire equation by 4 to simplify it: \(x^2 + y^2 = \frac{1}{4}\). This matches the standard form where \((h, k) = (0, 0)\) and \(r^2 = \frac{1}{4}\).

Step 3: Identify the Center of the Circle

From the simplified equation, the center is \((h, k) = (0, 0)\) since there are no \(h\) and \(k\) values that shift \(x\) or \(y\).

Step 4: Calculate the Radius

We have \(r^2 = \frac{1}{4}\). To find \(r\), take the square root of both sides: \(r = \sqrt{\frac{1}{4}} = \frac{1}{2}\).

Key Concepts

Standard Equation of a Circle

The standard equation of a circle helps us define the position and the size of the circle on a coordinate plane. It is expressed as \((x - h)^2 + (y - k)^2 = r^2\), where:

• \((h, k)\) represents the center of the circle, essentially moving the circle horizontally and vertically from the origin.
• \(r\) is the radius, indicating how far the circle extends from the center point.

Understanding this equation allows you to visualize the circle's position and figure out important details about it.
In problems like the original exercise, identifying this equation's form is the first step. By doing so, you can dissect and simplify any circle-related equation to pinpoint the circle's center and radius effectively.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is a crucial step in solving equations involving circles. In our original exercise, we began with the equation \(4x^2 + 4y^2 = 1\). The task was to simplify this expression to make it easier to analyze.
To do this, we divided every term by 4, transforming our equation into \(x^2 + y^2 = \frac{1}{4}\). This step was vital to match it with the standard form of a circle equation.

• Dividing helps to reduce coefficients and isolate variables.
• Simplification aligns the equation with recognizable mathematical forms, revealing key elements like the center and radius.

By simplifying effectively, we gain clearer insights into the structure of the equation, leading to more straightforward problem-solving.

Center of a Circle

Finding the center of a circle from its equation is key to understanding its location in a plane. The terms \((x-h)^2\) and \((y-k)^2\) in the standard form show where the center, \((h, k)\), is located.
For the exercise, after simplifying the equation to \(x^2 + y^2 = \frac{1}{4}\), we recognized no additional terms for \(x\) and \(y\), meaning the circle’s center is at \((0, 0)\), the origin.

• If the circle was shifted, \(h\) and \(k\) would appear in the equation.
• Identifying \(h\) and \(k\) helps in visualizing how far and in which direction the circle moves from the origin.

The center is a fundamental aspect because it acts as the pivot from which the radius of the circle extends.

Radius of a Circle

The radius of a circle describes its size by giving the distance from the center to any point on its edge. In the standard equation, \(r^2\) is the term representing the squared radius.
After simplifying \(x^2 + y^2 = \frac{1}{4}\), it becomes clear that \(r^2 = \frac{1}{4}\). Taking the square root of both sides, we find \(r = \sqrt{\frac{1}{4}} = \frac{1}{2}\).

• This operation transforms the squared radius back to its actual distance measure.
• The radius dictates the circle's size, which is crucial for drawing or analyzing it in problems.

Understanding the radius helps solve geometric problems and provides insights into the circle's properties in any given situation.