Problem 20 - Algebra for College Students

Question

For Problems \(15-32\), find the center and the length of a radius of each of the circles. $$ 5(x-3)^{2}+5(y-3)^{2}=30 $$ \((3,3) ; r=\sqrt{6}\)

Step-by-Step Solution

Answer

Center: (3,3); Radius: \(\sqrt{6}\)

Step 1: Analyze the Equation

The given equation is \( 5(x-3)^2 + 5(y-3)^2 = 30 \). This is a standard form equation for a circle: \( (x-h)^2 + (y-k)^2 = r^2 \). We need to transform the equation to match this form.

Step 2: Simplify the Equation

First, divide the entire equation by 5 to make the coefficients of \((x-3)^2\) and \((y-3)^2\) equal to 1. This results in: \( (x-3)^2 + (y-3)^2 = 6 \).

Step 3: Identify the Center

The center of the circle is given by \((h, k)\) in the equation \((x-h)^2 + (y-k)^2 = r^2\). Comparing this with \((x-3)^2 + (y-3)^2 = 6\), we see that \(h=3\) and \(k=3\). So, the center is \((3,3)\).

Step 4: Find the Radius

The right side of the equation \((x-3)^2 + (y-3)^2 = 6\) indicates \(r^2 = 6\). To find \(r\), we take the square root, giving \(r = \sqrt{6}\).

Key Concepts

Understanding the Center of a Circle

The center of a circle is a fundamental component to comprehend when dealing with the equation of a circle. Imagine the circle as a perfectly balanced wheel, and the center is like the axle of that wheel: everything revolves around it. In mathematical terms, if you have the equation \((x-h)^2 + (y-k)^2 = r^2\), the center is located at the coordinates \((h, k)\). This means:

• \(h\) is the x-coordinate of the center.
• \(k\) is the y-coordinate of the center.

For instance, in the equation \((x-3)^2 + (y-3)^2 = 6\), when you compare it to the standard circle equation, you can identify that the center is at \((3, 3)\). This is because \(h = 3\) and \(k = 3\). Understanding the center not only helps locate the circle on a coordinate plane but is also critical in defining the circle's unique position within the plane.

Determining the Radius of a Circle

The radius of a circle is the distance from its center to any point on the circle itself. You can think of it as the "arm's reach" of the circle. In the circle's equation \((x-h)^2 + (y-k)^2 = r^2\), \(r\) is not directly present, but \(r^2\) is on the right side of the equation. To find the radius, you follow these steps:

• Identify \(r^2\) from the equation.
• Calculate the square root of \(r^2\) to find \(r\).

For example, with the modified equation \((x-3)^2 + (y-3)^2 = 6\), we see that \(r^2 = 6\). Taking the square root, we find \(r = \sqrt{6}\). A circle's radius is vital for understanding the size of the circle and solving problems like finding the circumference or area.

Standard Form of a Circle Equation

The standard form of a circle's equation is a concise way of representing a circle using algebra. It looks like this: \((x-h)^2 + (y-k)^2 = r^2\). This form allows you to readily identify both the circle's center and its radius:

• The terms \((x-h)^2\) and \((y-k)^2\) involve shifting the whole structure along the x and y-axis to center it at \((h, k)\).
• The \(r^2\) on the right informs us about the circle's radius squared.

When given an equation not initially in this format, like \(5(x-3)^2 + 5(y-3)^2 = 30\), we simplify by dividing by 5 to align it with the standard form, resulting in \((x-3)^2 + (y-3)^2 = 6\). This transformation reveals the center as \((3, 3)\) and helps calculate the radius. It's crucial because it gives direct access to the key features of the circle, making problem-solving much more straightforward.