CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    The centre of the sphere passing through the origin and through the intersection points of the plane\[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\]with axes is:

    A)  \[\left( \frac{a}{2},0,0 \right)\]                 

    B)  \[\left( 0,\frac{a}{2},0 \right)\]

    C)  \[\left( 0,0,\frac{a}{2} \right)\] 

    D)         \[\left( \frac{a}{2},\,\frac{b}{2},\,0 \right)\]

    E)  \[\left( \frac{a}{2},\frac{a}{2},\frac{c}{2} \right)\]

    Correct Answer: E

    Solution :

    Let the equation of sphere be \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2vy+2wz+d=0\] It passes through origin \[\therefore \]  \[d=0\] Also it passes through (a, 0, 0), (0, b, 0), (0, 0, c) \[\Rightarrow \]            \[{{a}^{2}}+2ua=0\] \[\Rightarrow \]               \[{{a}^{2}}=-2ua\Rightarrow u=-\frac{a}{2}\] Similarly,   \[v=-\frac{b}{2},w=-\frac{c}{2}\] \[\therefore \]Centre \[(-\text{ }u,-\text{ }v,-w)=\left( \frac{a}{2},\frac{b}{2},\frac{c}{2} \right)\].


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