CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    The radius of any circle touching the lines \[3x-4y+5=0\]and\[6x-8y-9=0\]is:

    A)  1.9                                        

    B)  0.95

    C)  2.9                        

    D)         1.45

    E)  1.95

    Correct Answer: B

    Solution :

    Distance between the lines\[3x-4y+5=0\]and\[6x-8y-9=0\]is                 \[\left| \frac{5+\frac{9}{2}}{\sqrt{{{3}^{2}}+{{4}^{2}}}} \right|=\left( \frac{19}{10} \right)=1.9\] is the diameter of circle \[\therefore \]Radius of circle \[=\frac{1.9}{2}=0.95\]


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