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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2005

  • question_answer
    The radius of the circle  \[{{x}^{2}}+{{y}^{2}}+4x+6y+13=0\] is ;

    A)  \[\sqrt{26}\]                                     

    B)  \[\sqrt{13}\]

    C)  \[\sqrt{23}\]                                     

    D)  0

    Correct Answer: D

    Solution :

    Given equation is                 \[{{x}^{2}}+{{y}^{2}}+4x+6y+13=0\] or \[\left( {{x}^{2}}+4x+4 \right)+\left( {{y}^{2}}+6y+9 \right)+13=4+9\] or \[{{\left( x+2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=0\] \[\therefore \] radius of circle = 0.


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