Consider that points P such that the distance from P to ..?

Question

Consider that points P such that the distance from P to A(1,-2,1) is twice the distance from P to (3,-4,5). Show that the set of all such points is a sphere, and find its center and the radius

Answer 1

d(PA) = 2d(PB)
§ squaring...
d(PA)² = 4d(PB)²

(x-1)² + (y+2)² + (z-1)² = 4[(x-3)² + (y+4)² + (z-5)²]

x² - 2x + 1 + y² + 4y + 4 + z² - 2z + 1 = 4x² - 24x + 36 + 4y² + 32y + 64 + 4z² - 40z + 100
3x² - 22x + 3y² + 28y + 3z² - 38z + 194 = 0

x² - (22/3)x + y² + (28/3)y + z² - (38/3)z = -194/3

(x-11/3)² + (y + 14/3)² + (z - 19/3)² = 121/9 + 196/9 + 361/9 - 582/9 = 96/9

center( 11/3, -14/3 , 19/3)
radius: (4√6)/3

Answer 2

| P - A | = 2| P - B|

| P - A |^2 = 4| P - B|^2

(px -ax)^2 + etc. = 4(px - bx)^2 + etc.

px^2 -2pxax + ax^2 - 4px^2 + 2pxbx - bx^2 + etc = 0

-3px^2 + 2px(bx-ax) + ax^2 - bx^2 + etc. = 0

px^2 - (2/3)px(bx - ax) - (ax^2-bx^2)/3 + etc. = 0

(px - (1/3)(bx - ax))^2 - [(1/3)(bx - ax)]^2 - (ax^2-bx^2)/3 + etc. =0

px is really our field variable, e.g., x.

Thus the center of the sphere is

(1/3)[(bx -ax), (by -ay), (bz - az)]

With a radius of sqrt( [(1/3)(bx - ax)]^2 + (ax^2-bx^2)/3 + etc.)