10.029. The common chord of two intersecting circles is equal to $a$ and serves as a side of a regular inscribed triangle for one circle; and as a side of an inscribed square for the other. Determine the distance between the centers of the circles (consider two possible cases).
Solution.

1st case. By the condition $A B=a=A N=N M=M B=B D=A D, O_{1} E=\frac{a}{2}$, $\angle B O_{2} E=2 \angle E D B=60^{\circ}, O_{2} E=E B \cdot \operatorname{ctg} 60^{\circ}=\frac{a}{2 \sqrt{3}}$,

$$
O_{1} O_{2}=\frac{a}{2 \sqrt{3}}+\frac{a}{2}=\frac{a}{6}(3+\sqrt{3})(\text { fig. } 10.29, a) .
$$

2nd case. $O_{1} O_{2}=O_{2} E-E O_{1}, E O_{1}=\frac{a}{2}, E O_{2}=E B \cdot \operatorname{ctg} 60^{\circ}=\frac{a}{2 \sqrt{3}}$;

$$
O_{1} O_{2}=\frac{a}{2 \sqrt{3}}-\frac{a}{2}=\frac{a}{6}(3-\sqrt{3})(\text { fig. } 10.29, \text { b) }
$$

Answer: $a(3+\sqrt{3}) / 6, a(3-\sqrt{3}) / 6$.