Question

Hi can someone help me determine how to solve this ?
Consider the points P at which the distance to A (-1,5,3) is twice that of P to B (6,2, -2). Show these points on a sphere and determine its centre and radius.

Answer 1

@zepdrix Hi, You helped me for a similar problem earlier, Would it be possible to help me with this one please?

Answer 2

I also have the answer if needed, i just don't know how to get to the solution

Answer 3

Use the distance formula to get
\[
(x+1)^2+(y-5)^2+(z-3)^2=4
\left((x-6)^2+(y-2)^2+(z+2)^2\right)
\]

Answer 4

After expanding and simplifying you get
\[
3 x^2-50 x+3 y^2-6 y+3 z^2+22 z+141=0
\]
Which is a sphere

Answer 5

Alright

Answer 6

So can we find the radius with this ?

Answer 7

Complete the squares and finish it

Answer 8

Wait do i find the radius from this equation ? Or does this equation represent coordinates ?

Answer 9

\[
x^2-\frac{50 x}{3}+\left(\frac{50}{6}\right)^2-\left(\frac{
50}{6}\right)^2\\+y^2-2 y+4-4+\\z^2+\frac{22
z}{3}+\left(\frac{22}{6}\right)^2-\left(\frac{22}{6
}\right)^2+47=0
\]

Answer 10

I have to be honest .. I'm not sure what's going on

Answer 11

You can now read the center and the radius

Answer 12

The answer my teacehr gave was C(25/3, 1, -11/3) with radius = \[\sqrt{332}/3\]

Answer 13

So basically we just use the distance formula for the problem ?

Answer 14

This might be right, I may have made a mistake for the y part.
Try to find my mistake

Answer 15

But why did you write 4 in front of the distance formula ?

Answer 16

Yes, I did. I should have added 1 and subtracted 1 instead of adding and subtracting 4

Answer 17

This should have been
\[
x^2-\frac{50 x}{3}+\left(\frac{50}{6}\right)^2-\left(\frac{
50}{6}\right)^2\\+y^2-2 y+1-1+\\z^2+\frac{22
z}{3}+\left(\frac{22}{6}\right)^2-\left(\frac{22}{6
}\right)^2+47=0
\]

Answer 18

The final answer is
\[
\left(x-\frac{25}{3}\right)^2+(y-1)^2+\left(z+\frac{11}{3}\
\right)^2=\left(\frac{\sqrt{332}}{3}\right)^2
\]

Answer 19

Do you understand it now?

Answer 20

But why did you write 4 in front of the distance formula ?

Answer 21

I get the after parts of the problem, It's just the beginning that has me perplexed

Answer 22

2^2 =4

Answer 23

We computed the square of the distances. If s = 2t then
\[
s^2 = 4 t^2
\]

Answer 24

Ooohhh I get it now lol

Answer 25

So the problem is pretty simple

Answer 26

Yes it is

Answer 27

I understand it now .. Thanks:D

Answer 28

YW