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

  • question_answer
    The value of k so that \[{{x}^{2}}+{{y}^{2}}+kx+4y+2=0\] and \[2({{x}^{2}}+{{y}^{2}})-4x-3y+k=0\]cut orthogonally is  [Karnataka CET 2004]

    A)            \[\frac{10}{3}\]                       

    B)            \[\frac{-8}{3}\]

    C)            \[\frac{-10}{3}\]                      

    D)            \[\frac{8}{3}\]

    Correct Answer: C

    Solution :

               Here, \[{{g}_{1}}=\frac{k}{2},\,{{f}_{1}}=2,\,{{c}_{1}}=2\]                                            \[{{g}_{2}}=-1,\,{{f}_{2}}=\frac{-3}{4},\,{{c}_{2}}=\frac{k}{2}\]                    Condition for orthogonal intersection,                    Þ \[2({{g}_{1}}{{g}_{2}}+{{f}_{1}}{{f}_{2}})={{c}_{1}}+{{c}_{2}}\]                    Þ  \[2\,\left[ \frac{-k}{2}+\left( \frac{-3}{2} \right) \right]=2+\frac{k}{2}\]                    Þ \[-k-3=2+\frac{k}{2}\] Þ \[\frac{3k}{2}=-5\];  \[k=\frac{-10}{3}\].


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