#### Answer

$y$ is NOT a function of $x$.

#### Work Step by Step

Solve for $y$ in terms of $x$ to obtain:
$\begin{array}{ccc}
&x^2+(y-1)^2 &= &4
\\&(y-1)^2 &= &4-x^2
\\&\sqrt{(y-1)^2} &= &\pm\sqrt{4-x^2}
\\&y-1 &= &\pm\sqrt{4-x^2}
\\&y &= &1 \pm\sqrt{4-x^2}
\end{array}$
The last equation above will give two different values of $y$ for each value of $x$.
Thus, the equation does not represent $y$ as a function of $x$.