complex numbers equation i'm sturggling with the two following problems:
1) Find all complex numbers $z$ satisfying
$$|(1+i)z-2i|=2$$
2) Find all complex numbers $z$ satisfying
$$\left|\frac{z-3}{z-5}\right|=\frac{\sqrt{2}}{2}$$
Here is what i found :
1) : i write $z=a+ib$. Putting this is the equation i get that equation 1) implies that
$$a^2+b^2-2(a+b)=0$$
but i'm not able to simplify the solution more.
2) setting $z=a+ib$ and computing i get that the equation implies that $|z-3|=\frac{\sqrt{2}}{2}|z-5|$, hence $(a-3)^2+b^2=\frac{1}{2} (a-5)^2+b^2$. The problem is that it seems to lead to the fact that $a$ satisfies an equation of the second degree whch only admits non-real solutions... What did i do wrong?

For the first one, note that 
$$a^2+b^2-2(a+b)=0\iff (a-1)^2+(b-1)^2=2.$$
Try the second one in the similar way as above, but note that we have
$$(a-3)^2+b^2=\frac 12(a-5)^2+\color{red}{\frac 12}b^2.$$
You'll get

$(a-1)^2+b^2=8$

