Common chord AB=20cm Then, AM+MB=20cm AM=MB=10cm In △AO′M, (AO′)2=(AM)2+(MO′)2 ⇒(18)2=(10)2+(O′M)2 ⇒324−100=(O′M)2 ⇒MO′=√224 =√2×2×2×14×2 ⇒MO′=4√14cm In △AOM, (AO)2=(AM)2+(MO)2 ⇒(16)2=(10)2+(MO)2 ⇒256−100=(MO)2 ⇒MO=√156=√2×2×3×13 ⇒MO=2√39cm ∴ Distance between the centre OO′=O′M+MO =4√14+2√39cm