JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Sphere

  • question_answer
    The equation of the sphere concentric with the sphere \[2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-6x+2y-4z=1\]and double its radius is                                                               [Kerala (Engg.) 2005]

    A)                    \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-x+y-z=1\]

    B)                    \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-6x+2y-4z=1\]

    C)                    \[2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-6x+2y-4z-15=0\]

    D)                    \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+y-2z=1\]

    E)                    \[2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-6x+2y-4z-25=0\]

    Correct Answer: E

    Solution :

                       Equation of sphere is,                    \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+y-2z-\frac{1}{2}=0\]                    \ Centre of sphere \[=\left( \frac{3}{2},\frac{-1}{2},1 \right)\]                    Radius \[=\sqrt{\left( \frac{9}{4} \right)+\left( \frac{1}{4} \right)+1+\frac{1}{2}}=2\]                    Now, radius of required sphere = 4, which is concentric with the given sphere.                    Hence, equation of required sphere is,                    \[{{\left( x-\frac{3}{2} \right)}^{2}}+{{\left( y+\frac{1}{2} \right)}^{2}}+{{(z-1)}^{2}}=16\]                    i.e., \[2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-6x+2y-4z-25=0\].


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