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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    If the equation of the tangent to the circle \[{{x}^{2}}+{{y}^{2}}-2x+6y-6=0\]parallel to \[3x-4y+7=0\]is\[3x-4y+k=0,\]then the values of k are:

    A)  \[5,-35\]                             

    B)  \[-5,35\]                                                          

    C)  \[7,-32\]             

    D)         \[-7,32\]

    E)  \[3,-13\]

    Correct Answer: A

    Solution :

    The given equation of circle is \[{{x}^{2}}+{{y}^{2}}-2x+6y-6=0\] The centre and radius of circle are\[(1,-3)\]and 4 respectively. Length of perpendicular from\[(1,-3)\]to \[3x-4y+k=0\]is equal to radius 4. \[\therefore \]  \[\left| \frac{3+12+k}{\sqrt{9+16}} \right|=4\] \[\Rightarrow \]               \[15+k=\pm 20\] \[\Rightarrow \]               \[k=20-15=5\] Or           \[k=-20-15=-35\]


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