How do i prove this set has at most 2 elements? Let $w,\alpha\in\mathbb{C}$ and $\delta,\epsilon >0$ such that $(w,\delta)\neq (\alpha,\epsilon)$
Define $G=\{z\in\mathbb{C} : |z-\alpha|=\epsilon \text{ and } |z-w|=\delta\}$
How do i prove that $G$ has at most 2 elements?
This is geometrically trivial, but i'm not sure how to prove this precisely..
 A: Do you want an algebraic proof or a geometric one? Geometric, "intuitive" proofs can be precise too.
Here's the one you were probably thinking of. In the complex plane, the set of all $z \in \mathbb{C}$ so that $|z-\alpha| = \epsilon$ is a circle centered at $\alpha$ and with radius $\epsilon$; similarly, the set of all $z \in \mathbb{C}$ so that $|z-w| = \delta$ is a circle centered at $w$ and with radius $\delta.$ Therefore, the set $G$ is the points of intersection of the circles. Since $(w, \delta) \neq (\alpha, \epsilon),$ these circles are distinct, so they intersect at most twice, as requested. $\square$
An algebraic proof would involve writing $z = (x, y)$ and using definitions of magnitude, etc. It wouldn't be too pretty.
A: One way to prove this would be to describe $G$ as the following intersection
$$G = \{z \in \mathbb C : |z-\alpha| = \epsilon\} =: A \cap B:= \{z \in \mathbb C : |z-w|= δ\}.$$
If you then reformulate $A$ and $B$ as the set of $\mathbb C$-rational points of two curves (specifically circles) and use Bézout's Theorem (ensuring you account for the points at infinity), the result will follow.
A: Hint:
Start by taking conveniently $\alpha=0$ and $\epsilon=1$. If $z\in G$
then $z=e^{i\phi}$ for some $\phi\in[0,2\pi)$ and to be shown is
that there are at most two values for $\phi$ s.t. $\left|e^{i\phi}-w\right|=\delta$.
Setting $w=a+bi$ that comes to proving that $\left(\cos\phi-a\right)^{2}+\left(\sin\phi-b\right)^{2}=\delta^{2}$
has no more than two solutions in $[0,2\pi)$. Again it is convenient to solve this for a special case first, e.g. $w=a\in\mathbb R$. The more general cases can
be deduced from the special cases.
A: A geometric way to look at it which can be made algebraically precise is to notice that $3$ different points determine the center of a circle on which they lie:


*

*The set of points which have equal distance from $2$ points is a line.

*$2$ different lines intersect at most in $1$ point.

