A) \[a=b\pm 2c\]
B) \[a=b\pm \sqrt{2}c\]
C) \[a=b\pm c\]
D) None of these
Correct Answer: B
Solution :
The circles will touch each other, if the length of the common chord is zero\[i.e.,\] \[\therefore \] \[\sqrt{4{{c}^{2}}-2{{(a-b)}^{2}}}=0\] \[\Rightarrow \] \[2{{c}^{2}}={{(a-b)}^{2}}\] \[\Rightarrow \] \[a=b\pm \sqrt{2}c\]You need to login to perform this action.
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