Find the number of real roots for $x+\sqrt{a^2-x^2}=b$, $a>0$, $b>0$, as a function of $a$ and $b$ 
Given: (1) $x+\sqrt{a^2-x^2}=b$, $(a,b,x)\subset \mathbb R$, $a>0$, $b>0$.
Find: number of roots for (1), given possible values for $a$ and $b$.

This is a question from a book for the preparation for math contests.
It states as final answer: (a) 1 root if $b<a$; and (b) 2 roots if $a<b<a\sqrt{2}$.
I'm having difficulties on finding this answer. I don't know whether it is correct or perhaps I'm not finding the right approach.
I started moving $x$ in (1) to the left, to get
$$\sqrt{a^2-x^2}=b-x$$
Before proceeding with squaring both sides, I saved 2 needed conditions for checking the final solution (c1) $a^2-x^2\ge 0$ and (c2) $b-x\ge 0$. Then squaring both sides, we get:
$$a^2-x^2=b^2+x^2-2bx\Leftrightarrow 2x^2-2bx+(b^2-a^2)=0$$
with discriminant $\triangle$ defined by:
$$\triangle=4(2a^2-b^2)$$
From this it is easy to see that a condition for 2 roots is (c3) $\sqrt{2}a>b,$ and for 1 root is (c4) $\sqrt{2}a=b,$ as $a>0$ and $b>0$. Then I find the roots as $$x=\frac{2b\pm \sqrt{\triangle}}{4}=\frac{b\pm \sqrt{2a^2-b^2}}{2}$$
From this point, I can't see a way to reach the stated answer, if it is right.
Full solutions or helpful hints are welcome. Sorry if it is a duplicate.

You just forgot to check those conditions you mentioned. Indeed, you should check that for what values of $a$ and $b$ the conditions $-a\le x\le a$ and $b-x \ge 0$ are satisfied. The difficulty is to consider a few if then conditions.
If $b=\sqrt{2}a$ then $x=\frac{b}{2}$ is a potential solution. We should make sure that this solution satisfies our inequalities, that is, we should have $-a\le\frac{b}{2}\le a$ and $b-\frac{b}{2}\ge0$ which is equivalent to $b\le 2a$. But $b=\sqrt{2}a$ so this leads to $\sqrt{2}\le 2$ which is identically true. So if $b=\sqrt{2}a$ then the only answer is $x=\frac{b}{2}=\frac{\sqrt{2}}{2}a$.
If $b\gt\sqrt{2}a$ then $\Delta:=\sqrt{2a^2-b^2}<0$ and there is not real solution.
If $b\lt\sqrt{2}a$ then two potential answers will be $x_1=\frac{b}{2}+\frac{\sqrt{2a^2-b^2}}{2}$ and $x_2=\frac{b}{2}-\frac{\sqrt{2a^2-b^2}}{2}$. I  guess you can see what to do from here. Just put $x_1$ and $x_2$ into the inqualities and see what restrictions will be obtained on $a$ and $b$. Then check these restrictions with the condition $b\lt\sqrt{2}a$ to see under what conditions $x_1$ and $x_2$ can be a solution.
Below, you can see a plot of the function $f(x)=x+\sqrt{a^2-x^2}$ with $a=1$. You can see easily that for what values of $b$ the equation $f(x)=b$ has a solution and indeed exactly how many solutions.

