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 want to study the function
$$
f(x)=x+\sqrt{a^2-x^2}
$$
defined over $[-a,a]$. We have $f(-a)=-a$, $f(a)=a$; moreover
$$
f'(x)=1-\frac{x}{\sqrt{a^2-x^2}}
$$
for $x\in(-a,a)$. The derivative can only vanish where
$$
x=\sqrt{a^2-x^2}
$$
so $x\ge0$ and $x^2=a^2-x^2$, that is, $x=a/\sqrt{2}$. Note that
$$
f(a/\sqrt{2})=\frac{a}{\sqrt{2}}+\frac{a}{\sqrt{2}}=a\sqrt{2}
$$
and that $a/\sqrt{2}$ is a point of absolute maximum for $f$.
Therefore the equation $x+\sqrt{a^2-x^2}=b$ has


*

*no solution for $b<-a$

*one solution for $-a\le b\le a$

*two solutions for $a<b<a\sqrt{2}$

*one solution for $b=a\sqrt{2}$

*no solutions for $b>a\sqrt{2}$



An algebraic solution.
The equation $\sqrt{a^2-x^2}=b-x$ has solutions only for $x\le b$. Then you can square: $a^2-x^2=b^2-2bx+x^2$ or $2x^2-2bx+b^2-a^2$. The roots of this equation are
$$
\frac{b-\sqrt{2a^2-b^2}}{2}
\qquad\text{and}\qquad
\frac{b+\sqrt{2a^2-b^2}}{2}
$$
There is no solution for $2a^2-b^2<0$, that is, $b>a\sqrt{2}$.
There will be two solutions when $b<a\sqrt{2}$ and
$$
\frac{b+\sqrt{2a^2-b^2}}{2}\le b
$$
that is, 
$$
\sqrt{2a^2-b^2}\le b
$$
or $b\ge a$. Hence, $a\le b<a\sqrt{2}$.
There will be one solution when the largest root is greater than $b$, that is $b<a$, but also the smallest root is $\le b$, that is
$$
\frac{b-\sqrt{2a^2-b^2}}{2}\le b
$$
which is always satisfied when $b>0$.
When the discriminant is $0$, then there is a single solution.
