J & K CET Engineering J and K - CET Engineering Solved Paper-2011

  • question_answer The plane \[2x-2y+z+5=0\] is a tangent to the sphere \[{{(x-2)}^{2}}+{{(y-2)}^{2}}+{{(z-1)}^{2}}={{r}^{2}},\] if r equals

    A)  \[1\]               

    B)  \[2\]

    C) \[4\]              

    D)  None of these

    Correct Answer: B

    Solution :

    Given equation of sphere, \[{{(x-2)}^{2}}+{{(y-2)}^{2}}+{{(z-1)}^{2}}={{r}^{2}}\] Here, centre of sphere \[=(2,2,1)\] Let radius \[=r\] Equation of a plane is,  \[2x-2y+z+5=0\] Now, r = perpendicular distance on the plane from the centre of the sphere \[r=\frac{|2(2)-2(2)+(1)+5|}{\sqrt{4+4+1}}=\frac{|4-4+1+5|}{\sqrt{9}}\] \[r=\frac{6}{3}=2\]


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