Problem 22 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write the equation of the circle in standard form. Then identify its center and radius. $$\frac{9}{2} x^{2}+\frac{9}{2} y^{2}=1$$

Step-by-Step Solution

Answer

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

Step 1: Convert to Standard Form

Divide the entire equation by \(\frac{9}{2}\) to have it in a standard form: \[x^{2}+y^{2}=\left(\frac{2}{3}\right)^{2}\]

Step 2: Identify the Center

The center of the circle is at the origin \((0,0)\), as there are no constants added or subtracted from \(x\) or \(y\) in the equation.

Step 3: Determine the Radius

In the standard form equation, the radius is given by the square root of the constant on the right side of the equation. Since the constant here is \(\left(\frac{2}{3}\right)^{2}\), the radius of the circle is just the square root of that, \(\frac{2}{3}\).

Key Concepts

Standard Form

In geometry, the standard form of the equation of a circle makes it easy to see the circle's attributes. The standard form is written as \[(x - h)^2 + (y - k)^2 = r^2\]where

• \((h, k)\) represents the coordinates of the center of the circle, and
• \(r\) is the radius of the circle.

You form the standard equation by ensuring variables \(x\) and \(y\) are squared and grouped on the left, while a constant is on the right.
In this problem, the equation \(x^{2}+y^{2}=\left(\frac{2}{3}\right)^{2}\) is in standard form. This tells us the center is positioned at the origin, as there are no numbers added or subtracted to \(x\) and \(y\). The right side reflects the square of the radius. This concise setup simplifies identifying the center and radius.

Center of the Circle

The center of a circle is an essential feature of its equation in standard form. This point, defined by coordinates \((h, k)\), is where the circle is balanced. If the circle's equation is written as \[(x - h)^2 + (y - k)^2 = r^2\]then \(h\) and \(k\) are what you look for in the equation. When the standard form appears as \(x^2 + y^2 = r^2\), it means the circle is centered at the origin, which is the point \((0, 0)\).
In our current problem, since there are no terms like \(x - h\) or \(y - k\), it confirms that both \(h\) and \(k\) are zero.
This places the circle's center precisely at the origin, emphasizing its symmetrical positioning around this key point.

Radius of the Circle

The radius of a circle can be effortlessly read from its standard form equation. The radius, denoted by \(r\), is simply the square root of the constant on the right side of the equation \[(x - h)^2 + (y - k)^2 = r^2\].
In this task, the equation is given as \[x^2 + y^2 = \left(\frac{2}{3}\right)^2\].
Thus, the radius \(r\) becomes the square root of \(\left(\frac{2}{3}\right)^2\). Calculating it, we find:

• \(r = \frac{2}{3}\)

The radius represents the distance from the center to any point on the circle's boundary. With a radius of \(\frac{2}{3}\), it means that every point on this circle lies \(\frac{2}{3}\) units from the origin, maintaining perfect consistency in this distance.