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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2000

  • question_answer
    The    radius    of    the    circle \[{{x}^{2}}+{{y}^{2}}+4x-6y-12=0\]is:

    A)  9                                            

    B)  5

    C) \[-3\]                                    

    D) \[-6\]

    Correct Answer: B

    Solution :

    \[{{x}^{2}}+{{y}^{2}}+4x-6y-12=0\] \[g=2,f=-3,c=-12\] Radius \[=\sqrt{{{g}^{2}}+{{f}^{2}}+c}=\sqrt{4+9+12}=5\]


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