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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    The circle \[{{x}^{2}}+{{y}^{2}}-3x-4y+2=0\]cut \[x-\]axis at:

    A)  \[(2,0),(-3,0)\] 

    B) \[(3,0),(4,0)\]

    C)  \[(1,0),(-1,0)\] 

    D)  \[(1,0),(2,0)\]

    Correct Answer: D

    Solution :

    Equation of circle \[{{x}^{2}}+{{y}^{2}}-3x-4y+2=0\]and it cuts the \[x-\]axis, therefore \[y=0\] \[{{x}^{2}}+0-3x-0+2\Rightarrow {{x}^{2}}-3x+2=0\] \[{{x}^{2}}-2x-x+2=0\]                 \[\Rightarrow \]\[x(x-2)-1(x-2)=0\]                 \[\Rightarrow \]\[(x-2)(x-1)=0\Rightarrow x=1,2\] Therefore the points are \[(1,0),(2,0).\]


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