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  • Need explanation for: - Three Dimensional Geometry - JEE Main-2

The radius of the circle in which the sphere x^{2}+y^{2}+z^{2}+2x-2y-4z-19=0  is cut by the plane x+2y+2z+7=0  is

  • Option 1)

    2

  • Option 2)

    3

  • Option 3)

    4

  • Option 4)

    1

 

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 |}

 

-

 Center of sphere = (-1,1,2)

radius= \sqrt{1+1+4+19} = 5

Perpendicular distance = \frac{12}{3}= 4

Hence, r^{2}= 5^{2}-4^{2}= 3^{2}

r= 3

 


Option 1)

2

Incorrect option

Option 2)

3

Correct option

Option 3)

4

Incorrect option

Option 4)

1

Incorrect option

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