CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    The radius of the sphere passing through the points (0, 0, 0), (a, 0, 0), (0, b, O) and (0, 0, c) is:

    A)  \[\frac{1}{2}{{({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}^{\frac{1}{2}}}\]

    B)  \[{{({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}^{\frac{1}{2}}}\]

    C)  \[2{{({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}^{\frac{1}{2}}}\]

    D)  \[a+b+c\]

    E)  \[\frac{{{({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}^{\frac{1}{2}}}}{\sqrt{2}}\]

    Correct Answer: A

    Solution :

    Let the equation of sphere be \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2vy+2wz+d=0\] Since, it passes through (0,0,0), (o,0,0), (0,b,0) (0,0,c) \[\Rightarrow \] \[d=0,\text{ }{{a}^{2}}+2ua=0\] \[\Rightarrow \]\[2u=-\text{ }a\] \[\Rightarrow \]                        \[u=-a/2\] Similarly, \[v=-\text{ }b/2,\text{ }w=-\text{ }c/2\] \[\therefore \]radius \[=\frac{1}{2}\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]


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