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Find the plane of the intersection of `x^2+y^2+z^2+2x+2y+2=0` and `4x^2+4y^2+4z^2+4x+4y+4z-1=0.`

Text Solution

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The given spheres are
`" "x^(2)+y^(2)+z^(2)+2x+2y+2z+2=0" "`(i)
and `" "x^(2)+y^(2)+z^(2)+x+y+z -(1//4) = 0 " "` (ii)
The required plane is
`" "(2x-x)+(2y-y)+ (2z-z)+ 2+ (1)/(4) =0 `
or `" "4x+4y +4z+9=0`
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Knowledge Check

  • The circle x^2 + y^2 - 2x - 4y + 1 = 0 and x^2 + y^2 + 4x + 4y - 1 = 0

    A
    touches internally
    B
    touch externally
    C
    have 3x + 4y - 1 = 0 the common tangent at the point of contact
    D
    have 3x + 4y + 1 = 0 as the common tangent at the point of contact

  • The equation of the plane through the line of intersection of the planes 2x+y-z+5=0 and x+2y+3z=4 and perpendicular to the plane 5x+3y+6z=10 is -

    A
    `51x+15y-50z=173`
    B
    `5x-15y+50z+117=0`
    C
    `51x+15y-50z+173=0`
    D
    `63x-43y-50z+117=0`

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