CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    The values of\[\lambda \] so that the line\[3x-4y=\lambda \] touches\[{{x}^{2}}+{{y}^{2}}-4x-8y-5=0\]are:

    A)  \[-35,15\]          

    B)         \[3,-5\]

    C)  \[35,-15\]          

    D)         \[-3,5\]

    E)  20, 15

    Correct Answer: A

    Solution :

    The centre of circle is (2, 4). Radius\[=\sqrt{4+16+5}=5\] \[\therefore \]\[\bot \]distance of\[3x-4y-\lambda =0\] from (2, 4) = 5 \[\Rightarrow \]               \[\left| \frac{6-16-\lambda }{\sqrt{9+16}} \right|=5\] \[\Rightarrow \]               \[-10-\lambda =\pm 5\times 5=\pm 25\] \[\Rightarrow \]               \[\lambda =-10\pm 25\] \[\Rightarrow \]               \[\lambda =-35,15\]


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