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

  • question_answer
    The equation of the circle which passes through the point of intersection of circles \[{{x}^{2}}+{{y}^{2}}-8x-2y+7=0\] and \[{{x}^{2}}+{{y}^{2}}-4x+10y+8=0\] and having its centre on\[y\]-axis, will be

    A)            \[{{x}^{2}}+{{y}^{2}}+22x+9=0\]                                

    B)            \[{{x}^{2}}+{{y}^{2}}+22x-9=0\]

    C)            \[{{x}^{2}}+{{y}^{2}}+22y+9=0\]                                

    D)            \[{{x}^{2}}+{{y}^{2}}+22y-9=0\]

    Correct Answer: C

    Solution :

               Using\[{{S}_{1}}+\lambda {{S}_{2}}=0\], but its centre is on y-axis.                    i.e., \[-8-4\lambda =0\] or\[\lambda =-2\].                    Hence required equation is\[{{x}^{2}}+{{y}^{2}}+22y+9=0\].


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