Finding all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ Find all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ for all $x>0$
Attempt: Let $$\sqrt{9-(x-a)^2}> \sqrt{16-x^2}$$
So $$x^2-(x-a)^2>7$$
So $$a(2x-a)>7\Rightarrow a^2-2ax+7<0$$
could some help me how to find range of $a,$ Thanks
 A: 
The functions 
$g(x;a) = \sqrt{9-(x-a)^2}$ and 
$f(x)=\sqrt{16-x^2}$ 
are semicircles whose domains are $x \in [a-3, a+3]$  and $x \in [-4,4]$ respectively. Pictured above are $f(x), g(x;-1)$ and $g(x;1)$.
As the above picture suggests, $\sqrt{9-a^2+2ax-x^2} \le \sqrt{16-x^2}$ for all  $x \in [-1,1]$ when $a \in [-1,1]$.
If $a \in [-7 -1] \cup [1,7]$, then the domains of $f$ and $g$ will overlap and, at least for some $x$ in that overlap region, we will have 
$\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$
A: For the square roots to be defined, we need $|x|\leq 4$ and $|x-a|\leq 3$.
To solve a quadratic inequality, it's useful to find its roots. We have
$$a^2-2ax+7=0\iff a=\frac{2x\pm\sqrt{4x^2-28}}{2}=x\pm\sqrt{x^2-7}$$
In particular, if $|x|\leq\sqrt{7}$ the polynomial is always nonnegative in $a$.
If $\sqrt{7}<|x|\leq 4$, then the solution set will be all $a$ such that $|x-a|\leq 3$ and
\begin{align}&x-\sqrt{x^2-7}<a<x+\sqrt{x^2-7}\\
\iff &-\sqrt{x^2-7}<a-x<\sqrt{x^2-7}\\
\iff &|a-x|=|x-a|<\sqrt{x^2-7}\end{align}
Now, observe that if $|x|\leq 4$, then $x^2-7\leq 9$, so that $\sqrt{x^2-7}\leq 3$. In the end, all we need is:

$$\sqrt{7}<|x|\leq 4\,\,\text{ and }\,\,|x-a|<\sqrt{x^2-7}.$$

