91Ó°ÊÓ

In algebra, a circle equation defines all the points on a plane that are equidistant from a fixed point known as the center. The standard form of a circle equation is \[(x - h)^2 + (y - k)^2 = r^2\]where

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

Our given equation \(x^2 + (y-3)^2 = 9\) is already in this form. By comparing, we see the circle is centered at \((0, 3)\) with a radius of 3. Understanding this helps visualize the circle as a set of points forming a perfect circle on a graph.
This visualization is essential for comprehending why there could be two potential \(y\)-values for each \(x\)-value.

A function in algebra is a relationship that uniquely associates members of one set with members of another set. When we express \(y\) as a function of \(x\), every \(x\)-value should have a single corresponding \(y\)-value.
In the example problem, solving \(x^2 + (y-3)^2 = 9\) for \(y\) gave us two expressions: \(y = 3 + \sqrt{9 - x^2}\) and \(y = 3 - \sqrt{9 - x^2}\). This indicates two unique \(y\) values for each \(x\), 
suggesting that \(y\) isn’t a function of \(x\).
To check for functions: simply verify if any single \(x\)-value maps to more than one \(y\)-value. If it does, then \(y\) is not a function of \(x\).
Functions are often depicted graphically where vertical line tests can help identify them. If a vertical line intersects the curve more than once, it verifies that it is not a function.

The square root is a fundamental concept in algebra representing a number, which, when multiplied by itself, gives the original number. We denote the square root of \(a\) as \(\sqrt{a}\), and it is vital in solving equations like \(y-3 \) in our example.
When handling equations, square roots yield both positive and negative results: thus, \(\sqrt{a} = b\) and \(\sqrt{a} = -b\). This dual solution trait is why, in our problem, we ended with two equations for \(y\).
Particularly about circles, when applying the square root on terms like \(9 - x^2\), it's crucial to consider the domain. This term under the square root must be non-negative to provide real number results, i.e., \(9 - x^2 \geq 0\), which translates to \(-3 \leq x \leq 3\) on the number line. This domain consideration helps avoid non-real numbers, providing only feasible results for \(y\).