CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer The     radius     of     the     circle \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2y-4z-11=0\]and \[x+2y+2z-15=0\]is:

    A)  \[\sqrt{3}\]                                       

    B) \[\sqrt{5}\]                        

    C)  \[\sqrt{7}\]                       

    D)        \[3\]

    E)  \[\sqrt{2}\]

    Correct Answer: C

    Solution :

    Since,     the     centre     of     sphere \[{{x}^{2}}+{{y}^{2}}+\text{ }{{z}^{2}}-2y-4z-11=0\]is (0, 1, 2) and radius is 4. \[\therefore \]Distance of a plane\[x+2y+2z-15=0\] from (0, 1, 2) \[=\frac{|0+2+4-15|}{\sqrt{1+4+4}}\]                 \[=\frac{9}{3}=3\] Now, \[NP=\sqrt{O{{P}^{2}}-O{{N}^{2}}}\]                 \[=\sqrt{{{4}^{2}}-{{3}^{2}}}=\sqrt{16-9}=\sqrt{7}\] \[\therefore \]Radius of circle\[=\sqrt{7}\]


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