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JCECE Engineering JCECE Engineering Solved Paper-2011

  • question_answer
    If the circles\[{{(x-a)}^{2}}+{{(y-b)}^{2}}={{c}^{2}}\]and\[{{(x-b)}^{2}}+{{(y-a)}^{2}}={{c}^{2}}\]touch each other, then

    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\[i.e.,\] \[\therefore \]  \[\sqrt{4{{c}^{2}}-2{{(a-b)}^{2}}}=0\] \[\Rightarrow \]               \[2{{c}^{2}}={{(a-b)}^{2}}\] \[\Rightarrow \]               \[a=b\pm \sqrt{2}c\]


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