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Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    Radius of the circle \[\overset{\to }{\mathop{{{\mathbf{r}}^{\mathbf{2}}}}}\,+\overrightarrow{\mathbf{r}}(2\widehat{\mathbf{i}}-\widehat{\mathbf{j}}-4\widehat{\mathbf{k}})-19=0\] \[\overrightarrow{\mathbf{r}}\cdot (\widehat{\mathbf{i}}-2\widehat{\mathbf{j}}+2\widehat{\mathbf{k}})+8=0\], is

    A)  5                                            

    B)  4

    C)  3                            

    D)         2

    Correct Answer: B

    Solution :

    Given, circle is intersection of sphere. \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2x-2y-4z-19=0\] ... (i) and plane\[x-2y+2z+8=0\]                           ... (ii) Centre of sphere is (-1, 1, 2). \[P=\] length of the perpendicular from (-1, 1, 2) upon Eq. (ii)                 \[=\frac{-1-2+4+8}{\sqrt{1+4+4}}=\frac{9}{8}=3\] \[R=\]Radius of the sphere\[=\sqrt{1+1+4+19}=5\] Radius of the circle\[=\sqrt{{{R}^{2}}-{{p}^{2}}}=\sqrt{25-9}\]                 \[=\sqrt{16}=4\]


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