JCECE Engineering JCECE Engineering Solved Paper-2005

  • 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) \[1.45\]

    Correct Answer: B

    Solution :

    Key Idea: If two parallel lines touching the circle, then radius of a circle is half the distance between the tangents. Since, given tangent lines                 \[3x-4y+5=0\] and\[3x-4y-\frac{9}{2}=0\]are parallel. \[\therefore \]Distance between the lines\[=\frac{|{{a}_{1}}-{{a}_{2}}|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]                  \[=\frac{\left| 5+\frac{9}{2} \right|}{\sqrt{9+16}}=\frac{19}{10}\] \[\therefore \]Radius\[=\frac{1}{2}\times \frac{19}{10}\]                  \[=\frac{19}{20}=0.95\]


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