Is $x^2$ $+$ $y^2$ = $25$ a function? General Question: Is $x^2$ $+$ $y^2$ = $25$ a function?
If I input 2 different x's and it resulted in only 1 y value for each.
This is definitely a function right?
 A: No it is not. It is the equation of a circle with center at the origin and radius $5$.
$$
x^2 + y^2 = 5^2
$$

The curve interpreted as graph of a function $y = y(x)$ would result in a function which is multivalued for all but two $x$-arguments, $x = \pm 5$.
$$
y = \pm\sqrt{5^2 - x^2}
$$
A: This is a bit more complicated then everyone is saying.
A function has two parts: a domain and a rule.
Here you have only specified a rule. The rule is that you plug in $x$ and $y$ and must have $x^2+y^2=25$ be true.
The domain is important. For example, if the domain is only $x=-5$ and $x=5$, then you have a function since it is well defined (passes the vertical line test). If you include all $x$, this is not a function since it fails the vertical line test.
HOWEVER, you can even go so far as to say that this can be a function of two variables. Thus, the function would be as follows
$$f:\Bbb{R}\times\Bbb{R}\to\Bbb{R}\text{  where   }(x,y)\mapsto x^2+y^2$$
In this case, there is no 'vertical line test' per se. The function is totally valid, since it is not multivalued in these three dimensions.
To be clear, the difference is that, in the first case, the function is a function of one variable, $x$. Thus we have a problem because when $x=0$, for example, the function can be $y=\pm 5$, and we need only one value.
In the second case, the function is a function of two variables, $x$ and $y$. There is no issue because when we plug in $(x,y)$ into the 'rule' we always get exactly one output.
A: A function cannot have more than one image for every element in the domain. In this case, we are determing whether or not there is more than one $y$ value that satisifies the equation for a given $x$. What we want to do is determine whether or not this is the case for $x^2+y^2=25$, which is just a circle of radius 5. 
A good way to test this idea would be to draw the circle and see if it passes the vertical line test, (is there a place in the graph where a vertical line can intersect it twice), which is equivalent to one element in the domain (the elements along the x-axis) having two images in the codomain (elements along the y-axis).
Drawing a vertical line down the middle of the circle (or through any part besides where $x=\pm 5$),  you will see that this intersects it in two places, so there is a point which has two images for the same element in the domain. Therefore this is not a function.
A: No I think that becomes an equation. A function would be like f(x,y) = x^2 + y^2.
