解:⊙O1和⊙O2相交于A、B两点,则O1O2垂直平分线段AB;
又O1A⊥O2A,则:
O1O2=√(O1A^2+O2A^2)=√5;
O1A*O2A=O1O2*AC(=2S⊿O1AO2),即:
2*1=√5*AC,AC=(2√5)/5.故AB=2AC=(4√5)/5cm.