Problem 10 - Algebra 2

Question

Find the center and radius of the circle with the given equation. Then graph the circle. $$ \left(x+\frac{2}{3}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{8}{9} $$

Step-by-Step Solution

Answer

Center: \((-\frac{2}{3}, \frac{1}{2})\), Radius: \(\frac{2\sqrt{2}}{3}\).

Step 1: Identify the standard form of a circle equation

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

Step 2: Match given equation with the standard form

Compare the given equation \((x+\frac{2}{3})^2 + (y-\frac{1}{2})^2 = \frac{8}{9}\) to the standard form. Here, \((h,k) = (-\frac{2}{3}, \frac{1}{2})\), and \(r^2 = \frac{8}{9}\).

Step 3: Solve for radius

To find the radius \(r\), take the square root of \(r^2\). Thus, \[r = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}.\]

Step 4: Write the center and radius

The center of the circle is \((-\frac{2}{3}, \frac{1}{2})\) and the radius is \(\frac{2\sqrt{2}}{3}\).

Step 5: Graph the circle

Plot the center of the circle at \((-\frac{2}{3}, \frac{1}{2})\). Use the calculated radius \(\frac{2\sqrt{2}}{3}\) to draw the circle around this point, ensuring the circle extends equally in all directions from the center.

Key Concepts

Center of a Circle

The center of a circle is a crucial concept when working with circle equations. In a standard circle equation of the form \((x-h)^2 + (y-k)^2 = r^2\), the center is given by the coordinates \((h, k)\). These two numbers, \(h\) and \(k\), tell us where the middle point of the circle is located on a coordinate plane. This middle point is where you'll "anchor" your circle when you go to graph it.
To find the center from an equation like \((x+\frac{2}{3})^2 + (y-\frac{1}{2})^2 = \frac{8}{9}\), you compare it to the standard circle equation:\

• The \((x-h)\) part in the equation becomes \(x+\frac{2}{3}\).
• The \((y-k)\) part is \(y-\frac{1}{2}\).

Thus, \(h\) must be \(-\frac{2}{3}\), and \(k\) is \(\frac{1}{2}\).
This tells us the center of the circle is at \((-\frac{2}{3}, \frac{1}{2})\). Knowing the center is critical for graphing, as it's the point from which your circle will extend equally in all directions.

Radius of a Circle

Understanding the radius of a circle is equally as important as the center when working with circle equations. The radius is the distance from the center of the circle to any point on its circumference. It is always a positive number and is written as \(r\) in the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\).
To find the radius, we solve for \(r\) by taking the square root of the value on the right-hand side of the equation. In our example, the equation is \((x+\frac{2}{3})^2 + (y-\frac{1}{2})^2 = \frac{8}{9}\).

• The corresponding \(r^2\) part of the equation is \(\frac{8}{9}\).
• To get \(r\), you need to find \(\sqrt{\frac{8}{9}}\), which simplifies to \(\frac{\sqrt{8}}{3}\) or \(\frac{2\sqrt{2}}{3}\).

Knowing how to find the radius will help in drawing the circle accurately. It determines how "big" your circle is by indicating the exact scale from the center point.

Graphing Circles

Graphing circles is an integral part of understanding circle equations. Once you know the center and the radius, graphing a circle on a coordinate plane becomes straightforward. Begin by marking the center of your circle on the coordinate plane at the coordinate points \((-\frac{2}{3}, \frac{1}{2})\). This point is your reference for drawing the circle.
The next step involves using the radius \(\frac{2\sqrt{2}}{3}\) to draw the circle itself. This requires you to measure the distance from the center point outwards in all directions.
Remember:

• Plot four key points in the easy-to-measure cardinal directions: above, below, left, and right of the center.
• Use the radius distance \(\frac{2\sqrt{2}}{3}\) to determine how far these points are from the center.
• Once these points are plotted, draw a smooth curve connecting them to complete the circle.

Graphing circles accurately involves a steady hand and understanding the scale and measurements, but it becomes easier with practice. The visual outcome helps you easily see and contextualize the position and size of the circle with respect to the coordinate plane.