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The plane x+2y-z=4 cuts the sphere x^{2}+y^{2}+z^{2}-x+z-2=0  in a circle of radius

  • Option 1)

    1

  • Option 2)

    3

  • Option 3)

    \sqrt{2}

  • Option 4)

    2

 

Answers (1)

best_answer

As we learnt in 

Distance of a point from plane (Cartesian form) -

The length of perpendicular from P(x_{1},y_{1},z_{1}) to the plane

ax+by+cz+d= 0 is given by  \frac{\left [ ax_{1}+by_{1} +cz_{1}+d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}

 

-

 

Centre is \left(\frac{1}{2},0,-\frac{1}{2} \right )      r=\sqrt{\frac{5}{2}}

Perpendicular distance from centre=\frac{\left | \frac{1}{2}+\frac{1}{2}-4 \ \right |}{\sqrt{6}}=\sqrt{\frac{3}{2}}

Radius^{2}= \left ( \sqrt{\frac{5}{2}} \right )^{2}-\left ( \sqrt{\frac{3}{2}} \right )^{2}=1


Option 1)

1

Correct Option

Option 2)

3

Incorrect Option

Option 3)

\sqrt{2}

Incorrect Option

Option 4)

2

Incorrect Option

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