Find the real values of the parameter ‘a’ for which atleast one complex number z =x+iysatisfies both the equality| z- ai | = a+4 and the inequality | z-2 |

Hello Student,
Please find the answer to your question below [|z-ai| = a + 4]
[|z-2| < 1]
For these two to have atleast one solution, then both circles must intersect
[|x+iy-ai| = a+4]
[|x+i(y-a)| = a + 4]
[|x+i(y-a)|^{2} = (a + 4)^{2}]
[x^{2}+(y-a)^{2} = (a + 4)^{2}]
1st circle intersect the x – axis at 1 & 3.
2nd circle intersect x – axis
[x^{2}+(0-a)^{2} = (a + 4)^{2}]
[x^{2} + a^{2} = a^{2} + 16 + 8a]
[x^{2} = 16 + 8a]
16 + 8a = 1
so a=-15/8 and
if 16+8a=9
a=-7/8
so -15/8<a<-7/8

Dear student,
Can you please mention the complete inequality. Is it either |z-2|> 0, or |z-2|<0 or any other number. Once, you can provide that detail, we can colve it further.
Regards
Sumit