question_answer

Two circles of radii 10 cm and 8 cm intersect each other and the length of common chord is 12 cm. The distance between their centres is _____.

A) \[\sqrt{7}\,cm\]           

B)        \[3\sqrt{7}\,cm\]                     

C)        \[4\sqrt{7}\,cm\]         

D)        \[(8+2\sqrt{7})\,cm\]

Correct Answer: D

Solution :

M is the mid-point of AB, \[\therefore \]   \[AM=6\,cm\] \[AO'({{r}_{1}})=10cm,\] \[AO\,({{r}_{2}})=8\,cm\]             AB is perpendicular to OO', then             In  \[\Delta AOM,\,\,100=36+O{{M}^{2}}\] [using pythagoras theorem]             \[\Rightarrow \]\[OM=8\,cm;\] In   \[\Delta AMO',\,64=36+M{{O}^{2}}\] \[\Rightarrow \]  \[\sqrt{28}=MO'\,\,\,\Rightarrow \,\,\,2\sqrt{7}=MO'\] \[\therefore \]   \[OO'=(2\sqrt{7}+8)\,\,cm\]