VIT Engineering VIT Engineering Solved Paper-2008

  • question_answer
    The centre and radius of the sphere\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+3x-4z+1=0\]are

    A)  \[\left( -\frac{3}{2},\,0,\,-2 \right);\frac{\sqrt{21}}{2}\]

    B)  \[\left( \frac{3}{2},\,0,\,2 \right);\sqrt{21}\]

    C)  \[\left( -\frac{3}{2},\,0,\,2 \right);\frac{\sqrt{21}}{2}\]

    D)  \[\left( -\frac{3}{2},\,2,\,0 \right);\frac{21}{2}\]

    Correct Answer: C

    Solution :

    The given equation of sphere is \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+3x-4z+1=0\] On comparing this equation with general equation of sphere \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2vy+2wz+d=0,\] we get \[u=\frac{3}{2},\]\[v=0,\]\[w=-2\]and \[d=1\] \[\therefore \]Coordinates of centre of sphere \[=(-u,-v,-w)\] \[=\left( -\frac{3}{2},\,0,2 \right)\] and radius of sphere\[=\sqrt{{{u}^{2}}+{{v}^{2}}+{{\omega }^{2}}-d}\] \[=\sqrt{\frac{9}{4}+4-1}=\sqrt{\frac{9+12}{4}}\] \[=\frac{\sqrt{21}}{2}\]


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