Problem 29 - Algebra for College Students

Question

For Problems \(15-32\), find the center and the length of a radius of each of the circles. $$ x^{2}+y^{2}=8 \quad(0,0), r=2 \sqrt{2} $$

Step-by-Step Solution

Answer

Center: \((0,0)\), Radius: \(2\sqrt{2}\).

Step 1: Identifying the Standard Form of a Circle Equation

We start by recognizing the given equation of the circle, which is \( x^2 + y^2 = 8 \). In the standard form of the circle equation \((x-h)^2 + (y-k)^2 = r^2\), we compare it with the given equation.

Step 2: Determining the Center of the Circle

Since the equation is \( x^2 + y^2 = 8 \) and there are no linear terms in \( x \) and \( y \), it implies \( h = 0 \) and \( k = 0 \). Therefore, the center of the circle is at the origin \((0,0)\).

Step 3: Finding the Radius of the Circle

The equation \( x^2 + y^2 = 8 \) is equated to \( r^2 = 8 \), which means \( r = \sqrt{8} \). Simplifying \( \sqrt{8} \) gives us \( r = 2\sqrt{2} \).

Step 4: Conclusion

The analysis shows that the circle has its center at \((0,0)\) and a radius of \(2\sqrt{2}\).

Key Concepts

Center of a Circle

The center of a circle is a fixed point from which every other point on the circle is equidistant. It is an essential element when describing a circle geometrically. To find the center of a circle from its equation, we refer to the standard form of a circle equation: \((x-h)^2 + (y-k)^2 = r^2\). Here:

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

When an equation of the circle appears without any terms of \(x\) or \(y\) (e.g., \(x^2 + y^2 = 8\)), it indicates that the center coordinates \((h, k)\) are both zero. This means, in the example \(x^2 + y^2 = 8\), the circle's center is at

• Point \((0,0)\), known as the origin.

Understanding this helps us determine where the circle is positioned on a graph. It confirms that the center of our circle in this exercise is right at the center of the coordinate plane.

Radius of a Circle

The radius of a circle is a line segment connecting the center of the circle to any point on its boundary. It is consistent throughout and is often denoted by \(r\). Once we have the standard form of a circle, \((x-h)^2 + (y-k)^2 = r^2\), the radius can be determined by solving for \(r\) in \(r^2 = \,\text{some number}\).For the given circle's equation \(x^2 + y^2 = 8\), we have:

• \(r^2 = 8\)
• Taking the square root, \(r = \sqrt{8}\)

The next step is to simplify \(\sqrt{8}\) using factorization:

• \(\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}\)

Thus, the circle's radius is \(2\sqrt{2}\), giving a clear measure of how far from the center every point on this circle lies.

Standard Form of a Circle Equation

Understanding the equation of a circle begins with recognizing its standard form: \((x-h)^2 + (y-k)^2 = r^2\). This format allows us to instantly identify both the center and the radius, paving the way for easier graphing and analytical calculations.Key components to remember:

• \((h, k)\) is the circle's center point. It helps position the circle on the coordinate plane.
• \(r\), derived from \(r^2\), is the radius, indicating the circle's size.

In settings where certain terms (like \(x\) or \(y\)) aren't present, such as \(x^2 + y^2 = 8\), it signifies a circle centered at \((0,0)\) since \(h = 0\) and \(k = 0\).Translating a circle's equation into standard form enables simpler manipulations. It also directly displays the attributes of the circle, making it a fundamental skill in geometry and analytic geometry contexts.