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JEE Main & Advanced Mathematics Circle and System of Circles Question Bank System of circles

  • question_answer
    If the circle \[{{x}^{2}}+{{y}^{2}}+6x-2y+k=0\] bisects the circumference of the circle \[{{x}^{2}}+{{y}^{2}}+2x-6y-15=0,\] then k = [EAMCET 2003]

    A)            21  

    B)            - 21

    C)            23  

    D)   - 23

    Correct Answer: D

    Solution :

               \[2{{g}_{2}}({{g}_{1}}-{{g}_{2}})\,+2{{f}_{2}}({{f}_{1}}-{{f}_{2}})={{c}_{1}}-{{c}_{2}}\]                               \[2(1)\,\,(3-1)+2(-3)\,\,(-1+3)=k+15\]                    \[4-12=k+15\] or \[-8=k+15\,\Rightarrow \,k=-23\].


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