| 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] |
| \[AX=BX=\frac{AB}{2}=\frac{8}{2}=4\,\,cm\] |
| \[\therefore \] \[OX=\sqrt{O{{A}^{2}}-A{{X}^{2}}}=\sqrt{{{(5)}^{2}}-{{(4)}^{2}}}\] |
| \[=\sqrt{25-16}=3\] |
| \[\therefore \] \[OP=2\times OX=3\times 2=6\,\,cm\] |
You need to login to perform this action.
You will be redirected in
3 sec
