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JEE Main & Advanced Mathematics Circle and System of Circles Question Bank Chord of contact of tangent, Pole and Polar

  • question_answer
    If the circle \[{{x}^{2}}+{{y}^{2}}=4\]bisects the circumference of the circle \[{{x}^{2}}+{{y}^{2}}-2x+6y+a=0\], then a equals                    [RPET 1999]

    A)            4    

    B)            -4

    C)            16  

    D)            -16

    Correct Answer: C

    Solution :

               The common chord of given circles is                    \[2x-6y-4-a=0\]                                               .....(i)                    Since, x2 + y2 = 4 bisects the circumference of the circle x2 + y2 ? 2x + 6y + a = 0, therefore, (i) passes through the centre of second circle i.e., (1, ? 3).                    \ 2 + 18 ? 4 ?a = 0 Þ a = 16.


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