Open Ball in Euclidean Plane is Interior of Circle

Theorem
Let $\R^2$ be the real number plane with the usual (Euclidean) metric.

Let $x = \tuple {x_1, x_2} \in \R^2$ be a point in $\R^2$.

Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball at $x$.

Then $\map {B_\epsilon} x$ is the interior of the circle whose center is $x$ and whose radius is $\epsilon$.

Proof
Let $S = \map {B_\epsilon} x$ be an open $\epsilon$-ball at $x$.

Let $y = \tuple {y_1, y_2} \in \map {B_\epsilon} x$.

Then:

But from Equation of Circle:
 * $\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 = \epsilon^2$

is the equation of a circle whose center is $\tuple {x_1, x_2}$ and whose radius is $\epsilon$.

The result follows by definition of interior and Open Ball of Point Inside Open Ball.