question_answer

Two circles of same radius 5 cm, intersect each other at A and B. If AB = 8 cm, then the distance between the centres is                                                                                                             [SSC (CGL) 2013]

A) 10 cm              

B) 4 cm 

C) 6 cm    

D) 8 cm

Correct Answer: C

Solution :

On joining AO and AP.

In \[\Delta AOX,\]by Pythagoras theorem,

\[A{{O}^{2}}=A{{X}^{2}}+O{{X}^{2}}\]

\[\Rightarrow \]   \[{{(5)}^{2}}={{\left( \frac{AB}{2} \right)}^{2}}+{{(OX)}^{2}}\]

\[\Rightarrow \]   \[25={{\left( \frac{8}{2} \right)}^{2}}+{{(OX)}^{2}}\]

\[\Rightarrow \]   \[{{(OX)}^{2}}=25-16\]

\[\therefore \]      \[OX=\sqrt{9}=3\,\,cm\]

Similarly, in \[\Delta APX,\]\[PX=3\,\,cm\]

\[\therefore \]Distance between the centre

\[=OX+PX=3+3=6\,\,cm\]

Alternate Method

\[OA=AP=5\,\,cm\]        [radius]

\[AX=BX=\frac{AB}{2}=\frac{8}{2}=4\,\,cm\]

\[\therefore \]      \[OX=\sqrt{O{{A}^{2}}-A{{X}^{2}}}=\sqrt{{{(5)}^{2}}-{{(4)}^{2}}}\]

\[=\sqrt{25-16}=3\]

\[\therefore \]      \[OP=2\times OX=3\times 2=6\,\,cm\]