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let |=|z-2|)
then let 
which gives+i(y-1)|=|(x-2)+iy|)
^2+(y-1)^2}=\sqrt{(x-2)^2+y^2}\\\\(y-1)^2=y^2\\\\-2y+1=0\\\\y=\frac{1}{2})
ie=\frac{1}{2})
now this gives us a line for which
to find which side of the line gives|\leq|z-2|)
we substitute a point in say 
(testing for 
) which does not satisfy the inequality as 
hence the solution is that\geq\frac{1}{2})
which gives
ie
now this gives us a line for which
to find which side of the line gives
hence the solution is that