RAJASTHAN ­ PET Rajasthan PET Solved Paper-2007

  • question_answer The radius of the circle in which the sphere\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-x+z-2=0\]cuts by the plane \[x+2y-z=4,\]is

    A)  3               

    B)  1

    C)  2               

    D)  \[\sqrt{2}\]

    Correct Answer: B

    Solution :

     Centre of the sphere\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-z+z-2=0\]is\[\left( \frac{1}{2},0,-\frac{1}{2} \right)\]and radius\[=\sqrt{\frac{1}{4}+\frac{1}{4}+2}=\frac{\sqrt{10}}{2}\] Distance of plane from the centre of sphere \[=\frac{\left| \frac{1}{2}+\frac{1}{2}-4 \right|}{\sqrt{1+4+1}}=\frac{3}{\sqrt{6}}\] \[\therefore \]Radius of circle\[=\sqrt{\frac{10}{4}-\frac{9}{6}}=\sqrt{\frac{30-18}{12}}\] \[=1\]

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