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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

  • question_answer
    The chord of the circle \[{{x}^{2}}+{{y}^{2}}-4x=0\] which is bisected at \[(1,0)\] is perpendicular to the line

    A) \[y=x\]

    B) \[x+y=0\]

    C) \[x=1\]

    D) \[y=1\]        

    Correct Answer: D

    Solution :

    Given, equation of circle \[{{x}^{2}}+{{y}^{2}}-4x=0,\] bisect  by \[(1,0)\] Daygram                 Centre  of circle \[=(2,0)\]                 Radius of circle =2                 From figure, AB is the chord of circle which bisected by the point \[(1,0)\] and perpendicular to the line \[y=1.\] Because it is parallel to x-axis while chord is parallel to y-axis.


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