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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    If the two circles\[2{{x}^{2}}+2{{y}^{2}}-3x+6y+k=0\]and\[{{x}^{2}}+{{y}^{2}}-4x+10y+16=0\]cut orthogonally, then the value of k is:

    A)  41                         

    B)                         14

    C)         4                                            

    D)         0

    E)         2

    Correct Answer: C

    Solution :

    Both the circles cut orthogonally, if \[2\left( \frac{-3}{4} \right)(-2)+2.\frac{3}{2}.\frac{10}{2}=\frac{k}{2}+16\] \[\Rightarrow \]               \[3+15=\frac{5}{2}+16\] \[\Rightarrow \]               \[k=4\]


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