smc

The circles can be described in the cartesian plane as being centered at $(-1,0),(1,0)$ and $(0,\sqrt{3})$ with radius $r$ by the equations $x^2+(y-\sqrt{3}{)}^2=r^2$ $(x+1{)}^2+y^2=r^2$ $(x-1{)}^2+y^2=r^2$. Solving the first 2 equations gives $x=1-\sqrt{3}\cdot y$ which when substituted back in gives $y=\frac{\sqrt{3}\pm \sqrt{r^2-1}}{2}$. The larger root $y=\frac{\sqrt{3}+\sqrt{r^2-1}}{2}$ is the point B' described in the question. This root corresponds to $x=-\frac{1+\sqrt{3(r^2-1)}}{2}$. By symmetry across the y-axis the length of the line segment $B^{\prime }{C}^{\prime }$ is $1+\sqrt{3(r^2-1)}$ which is $\boxed{1+\sqrt{3(r^2-1)}}$.