Problem 21 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write the equation of the circle in standard form. Then identify its center and radius. $$\frac{4}{3} x^{2}+\frac{4}{3} y^{2}=1$$

Step-by-Step Solution

Answer

The standard form of the equation is \( (x - 0)^{2} + (y -0)^{2} = (\frac{\sqrt{3}}{2})^{2} \). The center is at the origin (0, 0) and the radius is \( \frac{\sqrt{3}}{2} \).

Step 1: Analyze the given equation

The given equation represents a circle and it is in the form of \( \frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}} = 1 \). The coefficients of \( x^{2} \) and \( y^{2} \) are already equal, so no further adjustment is needed.

Step 2: Write the standard form of the equation

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. However, in this case, since the equation doesn't have any h and k values, we know the center of the circle is (0,0).

Step 3: Identify the radius of the circle

In the given equation, we can extract the value for the radius directly. The standard form is \( \frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}} = 1 \), so for the given equation \( \frac{4}{3} x^{2}+\frac{4}{3} y^{2}=1 \), we can equate \( r^{2} = \frac{3}{4} \). Therefore, \( r = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \).

Key Concepts

Standard Form of a Circle

The standard form of a circle's equation is essential to understand and write circle equations. In mathematics, the standard form expresses the equation of a circle as \((x - h)^2 + (y - k)^2 = r^2\). Here, \(h\) and \(k\) represent the coordinates of the circle's center, and \(r\) represents its radius. This form makes it easy to identify the circle's properties right from the equation.
- If an equation of a circle shows no additional terms attached to \(x\) or \(y\), you can often conclude that the circle is centered at the origin \((0, 0)\).- If you have an equation like the one given \(\frac{4}{3} x^2 + \frac{4}{3} y^2 = 1\), you can see that the circles have both their \(x^2\) and \(y^2\) on one side and equal coefficients, indicating its form is akin to \(\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1\). This confirms that it retains the standard format without additional variables.

Center of a Circle

Identifying the center of a circle is a vital part of working with circle equations. The center \((h, k)\) is indicated right in the equation when it's in the standard form \((x - h)^2 + (y - k)^2 = r^2\). Here is what you should note:
- In our example equation \(\frac{4}{3} x^2 + \frac{4}{3} y^2 = 1\), you do not see any \(- h\) or \(- k\) terms attached to the \(x\) or \(y\). This indicates the circle's center is at the origin \((0, 0)\), as there are no shifts in the \(x\) or \(y\).
- Generally, if \((x - h)^2\) and \((y - k)^2\) have no additional constants being added or subtracted, the circle is centered at the origin. This concept makes locating the exact point of the circle's center straightforward.
Identifying the center explains how the circle is positioned within the coordinate plane, helping you predict and calculate its interactions with other geometric elements.

Radius of a Circle

The radius of a circle plays a crucial role in defining its size and dimensions. In a standard form equation \((x - h)^2 + (y - k)^2 = r^2\), \(r\) represents the radius of the circle, and being able to determine this from a given equation is fundamental. Let's consider the example equation provided.
- Our initial equation is \(\frac{4}{3} x^2 + \frac{4}{3} y^2 = 1\). Based on the previous standard circle form, we can rewrite this as \(\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1\), equating \(r^2\) to the reciprocal of the coefficients.- Since both \(x^2\) and \(y^2\) terms share the coefficient \(\frac{4}{3}\), you set \(r^2 = \frac{3}{4}\).
- To solve for \(r\), take the square root of \(\frac{3}{4}\). This gives \(r = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\).
Understanding the radius allows one to calculate various aspects of the circle, such as its circumference and area, furthering a comprehensive understanding of the circle's geometry.