Problem 9 - Algebra 2

Question

Find the center and radius of the circle with the given equation. Then graph the circle. $$ (x-4)^{2}+y^{2}=\frac{16}{25} $$

Step-by-Step Solution

Answer

The center is (4,0) and the radius is \(\frac{4}{5}\).

Step 1: Identify the Circle Equation Format

The given equation \((x-4)^2 + y^2 = \frac{16}{25}\) resembles the standard form of a circle equation: \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h,k)\) represents the circle's center, and \(r\) represents the radius.

Step 2: Extract the Center of the Circle

By comparing the equation \((x-4)^2 + y^2 = \frac{16}{25}\) to the standard form \((x-h)^2 + (y-k)^2 = r^2\), we identify \(h = 4\) and \(k = 0\). Thus, the center of the circle is \((4,0)\).

Step 3: Determine the Radius of the Circle

The right-hand side of the equation \(\frac{16}{25}\) represents \(r^2\), meaning the radius \(r\) is \(\sqrt{\frac{16}{25}}\). Calculating gives \(r = \frac{4}{5}\).

Step 4: Construct the Graph of the Circle

To graph the circle, plot the center at the point \((4, 0)\). Then, using a compass or drawing tool, draw a circle around the center with a radius of \(\frac{4}{5}\). Ensure the circle reaches approximately 0.8 units away from the center in all directions.

Key Concepts

Standard Form of a Circle Equation

To understand how to work with circles mathematically, knowing the standard form of a circle's equation is essential. This format is given by the equation \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h, k)\) denotes the center of the circle, and \(r\) represents the radius. This structure succinctly captures the circle's geometry.

• \( (x-h)^2 \) and \( (y-k)^2 \) components show the transformations along the x and y axes, centering the circle at \((h, k)\).
• The "+" sign indicates that the components form a perfect square.
• \( r^2 \) on the right side represents the squared radius, ensuring every point on the circle is \( r \) units away from the center.

Through this form, finding both the circle's radius and center becomes straightforward.

Center of a Circle

The center of a circle is a critical component in circle geometry, denoting the exact middle point around which all points on the perimeter equidistantly revolve. From the standard equation \((x-h)^2 + (y-k)^2 = r^2\), the center \((h, k)\) is derived directly from the expression by recognizing \(h\) and \(k\) as the constants modifying \(x\) and \(y\).
In the provided example equation \((x-4)^2 + y^2 = \frac{16}{25}\), comparing it to the standard form reveals:

• \( h = 4 \)
• \( k = 0 \)

Consequently, the center of this circle is \((4, 0)\).

Radius of a Circle

Understanding the radius within the context of the circle's equation is equally significant. The radius \(r\) is the distance from the center of the circle to any point on its boundary. In the equation \( (x-h)^2 + (y-k)^2 = r^2 \), \(r^2\) provides the square of the radius. To find \(r\), simply take the square root of \(r^2\).
For the equation \((x-4)^2 + y^2 = \frac{16}{25}\), \(r^2\) is determined to be \(\frac{16}{25}\), thus:

• \( r = \sqrt{\frac{16}{25}} = \frac{4}{5} \)

The radius here breaks down to \(0.8\), highlighting how far each part of the circle's perimeter reaches from its center.

Graphing Circles

Graphing circles involves a blend of mathematical understanding and precise drawing skills. After determining the circle's center and radius from its equation, begin plotting by marking the center point on a graph.
For the equation \((x-4)^2 + y^2 = \frac{16}{25}\):

• First, plot the center at \((4, 0)\).
• With the radius calculated as \(\frac{4}{5}\) or \(0.8\), use a compass or ensure precision by marking points \(0.8\) units away from the center in all directions.

Sketch the final circle smoothly, keeping its round symmetry intact. Proper graphing allows visualizing the balance and scale of the circle, further reinforcing the numerical calculations made.