JEE Main & Advanced Sample Paper JEE Main Sample Paper-24

  • question_answer
    If the circle \[{{x}^{2}}+{{y}^{2}}+4x+22y+c=0\] bisects  the circumference of the circle \[{{x}^{2}}+{{y}^{2}}-2x+8y-d=0\] then \[(c+b)\] equals

    A)  50                               

    B)  52

    C)  54                               

    D)  56

    Correct Answer: A

    Solution :

    \[{{s}_{1}}-{{s}_{2}}=0\] \[\Rightarrow \,6x+14y\,+c+d=0\] As (1,-4)lies on it, \[\Rightarrow \,6-56\,+c+d=0\] \[\Rightarrow \,c+d=50\]


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