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

  • question_answer
    The equation of a circle passing through points of intersection of the circles \[{{x}^{2}}+{{y}^{2}}+13x-3y=0\]and \[2{{x}^{2}}+2{{y}^{2}}+4x-7y-25=0\]and point (1, 1) is [RPET 1988, 89; IIT 1983]

    A)            \[4{{x}^{2}}+4{{y}^{2}}-30x-10y-25=0\]

    B)            \[4{{x}^{2}}+4{{y}^{2}}+30x-13y-25=0\]

    C)            \[4{{x}^{2}}+4{{y}^{2}}-17x-10y+25=0\]

    D)            None of these

    Correct Answer: B

    Solution :

               Required equation is                    \[({{x}^{2}}+{{y}^{2}}+13x-3y)+\lambda (2{{x}^{2}}+2{{y}^{2}}+4x-7y-25)=0\]                    which passes through (1, 1), so \[\lambda =\frac{1}{2}\].                    Hence required equation is                    \[4{{x}^{2}}+4{{y}^{2}}+30x-13y-25=0\].


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