Answer:
This is a sphere with a center [tex](\dfrac{8}{3},\dfrac{8}{3},\dfrac{8}{3} )[/tex] and radius [tex]\dfrac{4 \sqrt{3}}{3}[/tex]
Step-by-step explanation:
Given that: (x,y,z)
where;
d₁ is the distance from the origin = [tex]\sqrt{x^2+y^2+z^2}[/tex]
d₂ is the distance from (2,2,2) = [tex]\sqrt{(x-2)^2+(y-2)^2+(z-2)^2}[/tex]
Also;
d₁ = 2d₂
[tex]\sqrt{x^2 + y^2 +z^2 } = 2 \sqrt{(x-2)^2 +(y-z)^2+(z-2)^2}[/tex]
By squaring both sides;
[tex]x^2 + y^2 +z^2 = 2^2((x-2)^2 +(y-z)^2+(z-2)^2)[/tex]
[tex]x^2 + y^2 +z^2 = 4(x-2)^2 +4(y-z)^2+4(z-2)^2[/tex]
[tex]x^2 + y^2 +z^2 =4(x^2 -2x -2x +4) + 4(y^2 -2y -2y +4) + 4(z^2 -2z-2z+4)[/tex]
[tex]x^2 + y^2 +z^2 =4x^2 -16x +16 + 4y^2 -16y +16 + 4z^2 -16z+16[/tex]
Collect the like terms:
[tex]0 = 4x^2 -x^2 -16x +16 + 4y^2 -y^2 -16y +16 +4z^2 -z^2 -16z +16[/tex]
[tex]0 = 3x^2 -16x +16 + 3y^2 -16y +16 +3z^2 -16z +16[/tex]
Divide both sides by 3
[tex]\dfrac{0}{3} = \dfrac{3x^2 -16x +16}{3} + \dfrac{3y^2 -16y +16 }{3} + \dfrac{ +3z^2 -16z +16}{3}[/tex]
[tex]0 = {x^2 - \dfrac{16x}{3} +\dfrac{16}{3} + {y^2 - \dfrac{16y}{3} +\dfrac{16}{3}+{z^2 - \dfrac{16z}{3} +\dfrac{16}{3}[/tex]
Using the completing the square method;
[tex]3(\dfrac{160}{9}) -16 = {x^2 - \dfrac{16x}{3} +\dfrac{160}{9} + {y^2 - \dfrac{16y}{3} +\dfrac{160}{9}+{z^2 - \dfrac{16z}{3} +\dfrac{160}{9}[/tex]
[tex]\dfrac{16}{3}= (x - \dfrac{8}{3})^2+ (y - \dfrac{8}{3})^2+ (z - \dfrac{8}{3})^2[/tex]
∴
This is a sphere with a center [tex](\dfrac{8}{3},\dfrac{8}{3},\dfrac{8}{3} )[/tex] and ;
radius; [tex]\sqrt{\dfrac{16}{3}}[/tex]
[tex]= \dfrac{4 \sqrt{3}}{3}[/tex]
