Question

Show that the set of points that are twice as far from the origin as they are from (2, 2, 2) is a sphere. Find the center and radius of this sphere.

Answer 1

This is a sphere with a center
`((8)/(3),(8)/(3),(8)/(3) )` and radius
`(4 √(3))/(3)`

Explanation:

Given that: (x,y,z)

where;

d₁ is the distance from the origin =
`√(x^2+y^2+z^2)`

d₂ is the distance from (2,2,2) =
`√((x-2)^2+(y-2)^2+(z-2)^2)`

Also;

d₁ = 2d₂

`√(x^2 + y^2 +z^2 ) = 2 √((x-2)^2 +(y-z)^2+(z-2)^2)`

By squaring both sides;

`x^2 + y^2 +z^2 = 2^2((x-2)^2 +(y-z)^2+(z-2)^2)`

`x^2 + y^2 +z^2 = 4(x-2)^2 +4(y-z)^2+4(z-2)^2`

`x^2 + y^2 +z^2 =4(x^2 -2x -2x +4) + 4(y^2 -2y -2y +4) + 4(z^2 -2z-2z+4)`

`x^2 + y^2 +z^2 =4x^2 -16x +16 + 4y^2 -16y +16 + 4z^2 -16z+16`

Collect the like terms:

`0 = 4x^2 -x^2 -16x +16 + 4y^2 -y^2 -16y +16 +4z^2 -z^2 -16z +16`

`0 = 3x^2 -16x +16 + 3y^2 -16y +16 +3z^2 -16z +16`

Divide both sides by 3

`(0)/(3) = (3x^2 -16x +16)/(3) + (3y^2 -16y +16 )/(3) + ( +3z^2 -16z +16)/(3)`

`0 = {x^2 - (16x)/(3) +(16)/(3) + {y^2 - (16y)/(3) +(16)/(3)+{z^2 - (16z)/(3) +(16)/(3)`

Using the completing the square method;

`3((160)/(9)) -16 = {x^2 - (16x)/(3) +(160)/(9) + {y^2 - (16y)/(3) +(160)/(9)+{z^2 - (16z)/(3) +(160)/(9)`

`(16)/(3)= (x - (8)/(3))^2+ (y - (8)/(3))^2+ (z - (8)/(3))^2`

∴

This is a sphere with a center
`((8)/(3),(8)/(3),(8)/(3) )` and ;

radius;
`\sqrt{(16)/(3)}`

`= (4 √(3))/(3)`