Answer:
No because it contains points (3,4) and (3,-4). You cannot have a x assigned to more than one y-value if you want it to be a function.
Step-by-step explanation:
A function has one output per input.
If we are trying to determine if the given is a function of x, then x is the input.
However I can get two outputs from plugging in x=3.
![3^2+y^2=25](https://tex.z-dn.net/?f=3%5E2%2By%5E2%3D25)
![9+y^2=25](https://tex.z-dn.net/?f=9%2By%5E2%3D25)
Subtract 9 on both sides:
![y^2=25-9](https://tex.z-dn.net/?f=y%5E2%3D25-9)
![y^2=16](https://tex.z-dn.net/?f=y%5E2%3D16)
Take the square root of both sides:
![y=\pm \sqrt{16}](https://tex.z-dn.net/?f=y%3D%5Cpm%20%5Csqrt%7B16%7D)
.
So input x=3 yields y=4 and y=-4.
Since this input has more than one output then the given is not a function of x.
----Also!
If you graph the equation, it is a circle with radius 5 and center (0,0). So I could I plug in any number for x between -5 and 5 excluding -5 and 5 which would yield only one output each. Since plugging in either one gives:
![(\pm 5)^2+y^2=25](https://tex.z-dn.net/?f=%28%5Cpm%205%29%5E2%2By%5E2%3D25)
![25+y^2=25](https://tex.z-dn.net/?f=25%2By%5E2%3D25)
Subtract 25 on both sides:
![y^2=25-25](https://tex.z-dn.net/?f=y%5E2%3D25-25)
Simplify:
![y^2=0](https://tex.z-dn.net/?f=y%5E2%3D0)
There is only one value y such that when you square it gives you 0. That is 0.
x=5 only gives y=0 and x=-5 only gives y=0.
There is no circle, unless it is a circle with radius 0 which means it really wouldn't be a circle, that is a function.