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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • 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 :

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

    B)                  \[-\frac{8}{3}\]

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

    D)  \[\frac{8}{3}\]

    Correct Answer: B

    Solution :

    We have,                 \[{{x}^{2}}+{{y}^{2}}+kx+4y+2=0\]           ... (i)                 and                 \[2\left( {{x}^{2}}+{{y}^{2}} \right)-4x-3y+fc=0\]               ... (ii) from equation (i) \[\Rightarrow \]                 \[{{g}_{1}}=k/2,\,{{f}_{1}}=2\]and \[{{c}_{1}}=2\]and from equation (ii) \[\Rightarrow \]               \[{{g}_{2}}=-2,\,{{f}_{2}}=-\frac{3}{2},\,{{c}_{2}}=k\] Condition for two circles cut orthogonally is \[i.e.,\] \[2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}\] \[\Rightarrow \]               \[2\times \frac{k}{2}\times (-2)+2\times 2\times \left( -\frac{3}{2} \right)=2+k\] \[\Rightarrow \]                               \[-2\,k-6=2+k\] \[\Rightarrow \]                               \[-3k=8\] \[\Rightarrow \]                               \[k=-8/3\]


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