Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    The equation of the circle which passes through the origin and cuts orthogonally each of the circles\[{{x}^{2}}+{{y}^{2}}-6x+8=0\]and\[{{x}^{2}}+{{y}^{2}}-2x-2y=7\]is

    A) \[3{{x}^{2}}+3{{y}^{2}}-8x-13y=0\]

    B) \[3{{x}^{2}}+3{{y}^{2}}-8x+29y=0\]

    C) \[3{{x}^{2}}+3{{y}^{2}}+8x+29y=0\]

    D) \[3{{x}^{2}}+3{{y}^{2}}-8x-29y=0\]

    Correct Answer: B

    Solution :

    Let the required equation of circle be\[{{x}^{2}}+{{y}^{2}}+2gx+2fy=0\]. Since, the above circle cuts the given circles orthogonally. \[\therefore \]\[2(-3g)+2f(0)=8\] \[\Rightarrow \]  \[2g=-\frac{8}{3}\] and        \[-2g-2f=-7\] \[\Rightarrow \]               \[2f=+7+\frac{8}{3}=\frac{29}{3}\] \[\therefore \]Required equation of circle is                 \[{{x}^{2}}+{{y}^{2}}-\frac{8}{3}x+\frac{29}{3}y=0\] or            \[3{{x}^{2}}+3{{y}^{2}}-8x+29y=0\]


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