Everything Touches

$r$ denotes the radius and O is the center of small circle, C is the point where small circle tangents square. Let $\\ \angle ABO = \alpha, \angle CBO = \beta, AB = 1$

$\begin{aligned} AO &= 1 + r \\ BO &= 1 - r \\ \cos \alpha &= \sin \beta= \dfrac r{1-r} \\ AO^2 &= AB^2 + BO^2 - 2 \cdot AB \cdot BO \cos \alpha \\ (1+r)^2 &= 1 + (1-r)^2 - 2 \cdot 1 \cdot (1-r) \dfrac r{1-r} \\ r &= \dfrac 16 \end{aligned}$

Let $AB = 1$ and $A$ be the origin $(0,0)$ of the $xy$ -plane. Then the equations for the quadrants centered at $A$ and $B$ are:

$\begin{cases} Q_A: & x^2 + y^2 = 1 \\ Q_B: & (x-1)^2 + y^2 = 1 \end{cases}$

Let the center of the small circle be $C(x_0, y_0)$ and its radius $r$ . Then $C$ satisfies the following system of equations:

$\begin{cases} x_0^2 + y_0^2 = (1+r)^2 & ...(1) \\ (x_0-1)^2 + y_0^2 = (1-r)^2 & ...(2) \\ x_0 = 1 - r & ...(3) \end{cases}$

Then we have:

$\begin{aligned} (1)-(2): \quad x_0^2 - (x_0-1)^2 & = (1+r)^2 - (1-r)^2 \\ 2x_0 - 1 & = 4r & \small \blue{\text{From }(3): \ x_0 = 1 - r} \\ 2 - 2r - 1 & = 4r \\ \implies r & = \frac 16 \end{aligned}$

Therefore, $\dfrac {AB}r = \boxed 6$ .