A) \[2\sqrt{2}\]
B) 1
C) \[\sqrt{2}\]
D) 2
Correct Answer: D
Solution :
Now, let r be the equal radii of both the circles. APB is a right angled triangle. \[\therefore \]\[AB=\sqrt{2}r\] Also,\[AO=OB=\frac{r}{\sqrt{2}}\] In\[\Delta APO,{{\left( \frac{r}{\sqrt{2}} \right)}^{2}}+{{1}^{2}}={{r}^{2}}\Rightarrow r=\sqrt{2}\] \[\therefore \]\[AB=\sqrt{2}\times \sqrt{2}=2\]You need to login to perform this action.
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