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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If two chords having lengths\[{{a}^{2}}-1\]and\[3(a+1),\]where a is a constant of a circle bisect each other, then the radius of the circle is

    A)  6                                            

    B)  \[\frac{15}{2}\]

    C)  8                            

    D)         \[\frac{19}{2}\]

    E)  10

    Correct Answer: B

    Solution :

    Since, two chords bisect each other, it means both the chords passes through the centre of circle. \[\therefore \]Length of chords are equal. ie,         \[{{a}^{2}}-1=3(a+1)\] \[\Rightarrow \]               \[{{a}^{2}}-3a-4=0\] \[\Rightarrow \]               \[(a-4)(a+1)=0\] \[\Rightarrow \]          \[a=4\]           (\[\because \]\[a=-1\]is not possible) \[\therefore \]Radius of circle\[=\frac{{{a}^{2}}-1}{2}=\frac{16-1}{2}\]                 \[=\frac{15}{2}\]


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