Consider an inequality in form of circle equation (with zero coordinates):
$$y^2 + x^2 > 1$$
Expressing $y$ step-by-step (maybe I wrong somewhere?):
$$y^2 > 1 - x^2 (1)$$
$$y>\sqrt{(1-x^2)}\space ∨ \space y<-\sqrt{(1-x^2)} (2)$$
For $-1 ≤x ≤1$ it sure will be like this:
But what about other $x$? If $Y$ belongs to $R$, there should not be any $y$ for $|x|>1$.
But, Wolfram Alpha, for some reason plots for all $x$:
Is it wrong, and it should be only from $-1$ to $1$?
Maybe I expressed $y$ wrong?
First of all, the logical connective is $\vee$ and not $\wedge$; notice that there exists no real number $y$ such that $y>\sqrt{1-x^2}$ and $y<-\sqrt{1-x^2}$, so that inequality describes the empty set.
About your question, the inequalities: $$y^2>1-x^2$$ and $$\left(y>\sqrt{1-x^2}\right) \vee \left(y<-\sqrt{1-x^2}\right)$$ are not equivalent. This is because the square root function is defined only for nonnegative values, hence in the very moment you introduce the square root you're imposing $-1 \le x \le 1$. But your first inequality $y^2>1-x^2$ has no constraint on $x$, hence they are not equivalent. A way to solve this problem is to split the sets where $x$ belongs to, and then take the union. You notice that when $x> 1$ or $x<-1$, we have $1-x^2<0$ and so the inequality $y^2>1-x^2$ is true for all $y\in\mathbb{R}$ because the left hand side is nonnegative and the right hand side is negative.
When $-1 \le x \le 1$, we have $1-x^2 \ge 0$ and so we can take the square root both sides of $y^2>1-x^2$, obtaining the result you've already obtained. Putting those results together, you have the exterior of the open circle of radius $1$ and center in the origin.