Differential equation study: $y' = \frac{y}{x^2+y^2-1}$. Proof review. as told in the title I'm trying to study the following ODE:
$$y'=\frac{y}{x^2+y^2-1} \qquad y(0)=c \in (0,1)$$
Let $y$ be a maximal solution and let $(a,b)$ be the interval in which it's defined.

I'm asked to prove the following:
  
  
*
  
*$y>0$ in $(a,b)$;
  
*$y$ is decreasing in $(a,b)$;
  
*$-1<a<0$ and $\exists \ l=\lim_{x\to a}y(x)$ satisfying $a^2+l^2=1$;
  
*$b=1$ and $\lim_{x\to b}y(x)=0$.
  

That's my solution (I'm asking for a proof-review):
We first notice that $y=0$ is a stationary solution so, by uniqueness theorem, our solution $y(x)$ must not touch the above-mentioned line. So, as we start in $c > 0$, we must stay above $0$. And this answers the first question.
By studying the sign of $y'$ we have:
$$y'<0 \iff y>0 \ \wedge x^2+y^2<1$$
and so we answer the second point (I'm not considering outside cases because I don't have interests in them for the requests).
To prove the third point we notice that the derivative $y'$ explodes if $x^2+y^2=1$ and so, while not satisfying this, the solution is defined and it must remain inside the upper part of the circumference. So clearly we have $a \in (-1,0)$ and $b \in (0,1)$. Knowing that the solution is monotonically decreasing in $(a,b)$ we also know that exists $\lim_{x\to a} y(x)=l \in \mathbb{R}$, in particular we have $l \in (c,1)$. This limit must satisfy $a^2+l^2=1$ or the solution can be extended at the left of $a$ as the derivative is bounded. But this is absurd as we said that $(a,b)$ is the maximal interval in which the solution is defined.
To answer the fourth question I tried to find a super-solution like $v=\sqrt{1-(1-\epsilon)x^2}$ or $v=\sqrt{1-(1+\epsilon)x^2}$ but I am not really confident about the procedure. I'd like to say that $b=1$ and $\lim_{x\to b}y(x)=0$ it's the only possibility as the solution always stays under $y=\sqrt{1-x^2}$ when $x>0$ and must not trespass $0$, but I don't really know how to perfectly formalize it. 
Thanks in advance!

I see nothing to say about 1-3. For 4 one can analyze the equation to pieces by first changing the independent parameter to $t$ so that
\begin{align}
\dot y(t)&=-y,&y(0)&=c,\\
\dot x(t)&=1-x^2-y^2,&x(0)&=0,
\end{align}
and further parametrizing the second, Riccati equation via $x=\frac{\dot u}{u}$ to get the partially solved system
\begin{align}
y(t)&=ce^{-t}\\
u''(t)&=(1-c^2e^{-2t})u(t),~~~u(0)=1,~u'(0)=0
\end{align}
The solution for $u$ exists on the whole of $[0,\infty)$, is positive, increasing and convex there. Thus it has no roots that would translate into a singularity for $x(t)$. From the consideration of the whole system if follows that $y(t)\to 0$ and thus $x(t)\to 1$ for $t\to\infty$.
In a little more detail, the asymptotic behavior of $u(t)$ is given by $u(t)=Ae^t+Be^{-t}$ so that $x(t)\to 1$ for $t\to\infty$. Which means that the curve $(x(t),y(t))$ starts at $(c,0)$ and ends at $(1,0)$.
