How to calculate the circumference of a circle, excluding the part intersecting another circle? 
I have one circle of $11.63$ inch bricks that is $21.25$ ft in radius (green in the picture). I already purchased the bricks for this circle already.
I need to add another circle to the side (marked in red) of bricks. The size is scaled by the golden ratio of $1.618$, or $13.133$ ft in radius. The center of this red circle falls directly on the edge of the green circle.
I need to purchase bricks to form the second (red) circle, but I don't need bricks to go inside the green circle.
How do I find the circumference of circle B, minus the part intersected with circle A, so I can calculate how many bricks to buy?
 A: 
The $36^{\circ}-72^{\circ}-72^{\circ}$ $\triangle ABC$ is called the golden triangle as its sides are $a/\phi,a,a$.
Desired red circumference is
$$2\pi\cdot \frac{a}{\phi} \cdot\left(1-\frac{2\cdot72}{360}\right)$$
$$=2\pi\cdot \frac{a}{\phi}\cdot \frac{3}{5}$$
$$=\frac{6\pi}{5}\cdot \frac{a}{\phi} $$
where $a=21.25$
A: One possibility is to think about the intersection of the two circles. Assuming that the red circle is centered at $(0,0)$, its ''northeast'' quadrant is described by $y=\sqrt{r^2-x^2}$ (where $r$ is its radius). The green circle is centered at $(R,0)$, $R$ being its radius, and its ''northwest'' quadrant is described by $y=\sqrt{R^2-(x-R)^2}$. If you solve
$$\sqrt{r^2-x^2}=\sqrt{R^2-(x-R)^2}$$
you obtain the points at which the circles intersect: $\left(\frac{r^2}{2R},r\sqrt{1-\frac{r^2}{4R^2}}\right)$  and, by symmetry, $\left(\frac{r^2}{2R},-r\sqrt{1-\frac{r^2}{4R^2}}\right)$. These two points define an arc that spans $2\arctan\frac{r\sqrt{1-\frac{r^2}{4R^2}}}{\frac{r^2}{2R}}$ of the $360º$.
