<s>[INST] Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle So why is it a function, even though for example $x = 8$; you'll have $y = +2$ and $y = -2$. It'll fail the vertical line test. But every textbook considers it as a function. Did I misunderstand something?
Edit:
Wait how come $y = \sqrt{4 - x^2}$ is a function too when it can be transformed into
$$
y = \sqrt{4-x^2}
$$
$$ y^2 = 4- x^2$$
$$ x^2 + y^2 = 4 $$
$$ \frac{x^2}{4} + \frac{y^2}{4} = 1$$
which is an equation of a circle and not a function
 [/INST] For the followup question:
$y = \sqrt{x}$ and $y^2 = x$ are not equivalent statements.
If $y = \sqrt{x}$ then $y^2 = x$ but not the other way around. Hence your equations $$y = \sqrt{4-x^2}$$ and $$x^2 + y^2 = 4$$ may very well have different sets of solutions, and they do. For the first of these equations, $y$ is automatically positive (because of the other discussions about the square root). The solutions of $y = \sqrt{4-x^2}$ form a semi-circle, the part of the circle $x^2 + y^2 = 4$ which correspond to points where $y \ge 0$.
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