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

Answer:

This is a sphere with a center  $(\dfrac{8}{3},\dfrac{8}{3},\dfrac{8}{3} )$  and radius $\dfrac{4 \sqrt{3}}{3}$

Step-by-step explanation:

Given that: (x,y,z)

where;

d₁ is the distance from the origin = $\sqrt{x^2+y^2+z^2}$

d₂ is the distance from (2,2,2) = $\sqrt{(x-2)^2+(y-2)^2+(z-2)^2}$

Also;

d₁ = 2d₂

$\sqrt{x^2 + y^2 +z^2 } = 2 \sqrt{(x-2)^2 +(y-z)^2+(z-2)^2}$

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

$\dfrac{0}{3} = \dfrac{3x^2 -16x +16}{3} + \dfrac{3y^2 -16y +16 }{3} + \dfrac{ +3z^2 -16z +16}{3}$

$0 = {x^2 - \dfrac{16x}{3} +\dfrac{16}{3} + {y^2 - \dfrac{16y}{3} +\dfrac{16}{3}+{z^2 - \dfrac{16z}{3} +\dfrac{16}{3}$

Using the completing the square method;

$3(\dfrac{160}{9}) -16 = {x^2 - \dfrac{16x}{3} +\dfrac{160}{9} + {y^2 - \dfrac{16y}{3} +\dfrac{160}{9}+{z^2 - \dfrac{16z}{3} +\dfrac{160}{9}$

$\dfrac{16}{3}= (x - \dfrac{8}{3})^2+ (y - \dfrac{8}{3})^2+ (z - \dfrac{8}{3})^2$

∴

This is a sphere with a center  $(\dfrac{8}{3},\dfrac{8}{3},\dfrac{8}{3} )$  and ;

radius; $\sqrt{\dfrac{16}{3}}$

$= \dfrac{4 \sqrt{3}}{3}$