$AC = 5$ and $BC = 2$. Let $H$ and $K$ be the foot of the perpendicular bisectors of the first and the second triangle respectively, and $C_{1}$, $C_{2}$ their circumcenters. So $C_{1}H = \frac{5}{6}\sqrt{3}$ and $C_{2}K = \frac{\sqrt{3}}{3}$. Then $C_{1}C_{2} = \sqrt{13}$.

if centeres of triangle with base AC and CB are M and N respectively,then we can seeangle MCN =120.so using cosine rule on MCN we get MN=13^(1/2)

Let $BCD, CAE, ABF$ be equilateral triangles with $AB$ separating $F$ of $DE$. From Torricelli to "triangle $ABC$ we get $AD=BE=CF$. Find $CF$ by Stewart for $\Delta ABF$, see that $MN= \frac{AD}{\sqrt{3}}$ ($M, N$ as before). Best regards, sunken rock