<s>[INST] Complex number z satisfies both the inequality $|z-ai|=a+4$ and the inequality $|z-2|<1$ The number of integral values of $a$ for which at least one complex number z satisfies both the inequality $|z-ai|=a+4$ and the inequality $|z-2|<1$.
I supposed $z=x+iy$ and put in both equations,
$|x+iy-ai|=a+4\Rightarrow \sqrt{x^2+(y-a)^2}=a+4$$\Rightarrow x^2+y^2-2ya+a^2=a^2+8a+16...(1)$
$|x+iy-2|<1\Rightarrow \sqrt{(x-2)^2+y^2}<1.....(2) $
$\Rightarrow (x-2)^2+y^2<1$
$\Rightarrow x^2+y^2-4x+4<1......(2)$
How will i solve further to get integer values of $a$? 
 [/INST] HINT....Assuming $a$ is real, for a geometrical interpretation: the first equation represent the point on a circle centre $(0,a)$ and radius $a+4$, and the second one represents the points inside the circle centre$(2,0)$, radius $1$
</s>