<s>[INST] 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
 [/INST] 
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 
$\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$
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