To determine whether the equation $x^2 + (y - 2)^2 = 5$ defines $y$ as a function of $x$, we need to check if each value of $x$ corresponds to exactly one value of $y$.

Step-by-Step Analysis:

1. Rewrite the equation:
   The given equation is: $x^2 + (y - 2)^2 = 5$ This equation represents a circle with the center at $(0, 2)$ and radius $\sqrt{5}$.

2. Determine if $y$ is a function of $x$:
   For $y$ to be a function of $x$, each value of $x$ should correspond to only one value of $y$. However, in a circle, a single value of $x$ can correspond to two different values of $y$.

   For example, consider the equation when $x = 0$: $0^2 + (y - 2)^2 = 5 \implies (y - 2)^2 = 5.$ Solving for $y$: $y - 2 = \pm \sqrt{5}, \quad \text{so} \quad y = 2 + \sqrt{5} \quad \text{or} \quad y = 2 - \sqrt{5}.$ Here, for $x = 0$, there are two possible values for $y$.

3. Conclusion:
   Since one value of $x$ (like $x = 0$) corresponds to two different values of $y$, the equation does not define $y$ as a function of $x$.

Additional Explanation:

• A function $y = f(x)$ must pass the vertical line test: A vertical line drawn at any $x$-value on its graph must intersect the graph at most once. The equation $x^2 + (y - 2)^2 = 5$ represents a circle, which fails this test.

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Related Questions:

1. What is the domain and range of a circle defined by $x^2 + (y - 2)^2 = 5$?
2. How can we graph the equation $x^2 + (y - 2)^2 = 5$ and visualize why it isn't a function?
3. What is the vertical line test, and how does it determine if an equation is a function?
4. How do you solve for $y$ explicitly in the equation $x^2 + (y - 2)^2 = 5$?
5. What is the difference between a function and a relation in mathematics?

Tip: Remember that for a relation to be a function, every input (x-value) must correspond to exactly one output (y-value).