For any $ r \gt 0$ let $M=\lbrace z\in \mathbb C; |z| = 1 , |z - 3i| = r\rbrace$ and $A = \lbrace r \gt 0; |M| = 1 \rbrace$. Find the sum of the elements of $A$.
I tried to understand what I need to find. Here is what I thought:
$M$ is the set of the complex numbers that satisfy those two equalities. So any number $z = a + ib$ it needs to satisfy:
$(1) \sqrt{a^2+b^2} = 1$
$(2) \sqrt{a^2+ (b-3)^2} =r, \, r\gt 0$
Then $A$ is the set of the complex number for which the set $M$ contains just a number.
Did I get this right so far? Next I tried to find $a$ and $b$ by solving $(1)$ and $(2)$. I substituted $a^2 = 1 - b^2$ in the second equation and got:
$\sqrt{10-6b} = r, \, r\gt 0$
Now I have no clue what to do or even if I am on the right path. Help me a bit, please.
Hint: The Cartesian equation of $|z|=1$ is $x^2+y^2=1$. The Cartesian equation of $|z-3i|=r$ is $x^2+(y-3)^2=r^2$. Draw a sketch of these two circles. Notice there are only two possible values of $r$ that ensure the circles intersect in a single point. Can you see what they are?