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

  • question_answer
    The points of intersection of the circles \[{{x}^{2}}+{{y}^{2}}=25\]and \[{{x}^{2}}+{{y}^{2}}-8x+7=0\]are                                      [MP PET 1988]

    A)            (4, 3) and (4, -3)                     

    B)            (4, -3) and (-4, -3)

    C)            (-4, 3) and (4, 3)                     

    D)            (4, 3) and (3, 4)

    Correct Answer: A

    Solution :

                \[{{x}^{2}}+{{y}^{2}}=25\]                                                     .?. (i)                    \[{{x}^{2}}+{{y}^{2}}-8x+7=0\]                                                                     .?. (ii)                    From (i) and (ii), we get                    \[8x=7+25\]or\[x=4\] and for \[x=4\], we get \[y=\pm 3\].                    Hence points of intersection are (4, 3), (4, -3).


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