1. ## Sphere equation (impossible?)

Hi guys,

I just started in on my first math class that deals heavily with three dimensions, and already I'm stumped. The inability to draw clear graphs by hand is crippling!

So here's the problem that's got me. It's really not Calculus, but I didn't know where to post this, and since the class I'm taking is Calc 3, I figured I might as well post it here... sorry if it's the wrong place.

Find the equation of the sphere with points P such that the distance from P to A is twice the distance from P to B.
A(-2,5,2) B(5,2,-3)
So... if I understand correctly then EVERY point on the sphere must satisfy the requirement of having twice the distance from point A as from point B.
Since the whole three-dimensional thing is still so foreign to me, I figured I could get a better understanding by reducing it to 2 dimensions. So I took the cross-section of the plane containing the sphere's center and points A and B, and looked at the circle. I then applied a trial-and-error approach and made the circle centered on the origin with radius 1, then tried to come up with a configuration for points A and B to satisfy the original problem.
I wasn't able to do so.

It seems like the solution to the given condition would be something like an ellipsoid or hyperboloid... not a sphere. Am I missing something, or is this problem maybe just badly worded? Perhaps my prof meant "solid" instead of "sphere"?

Thanks!

2. Originally Posted by Fox2013
Hi guys,

I just started in on my first math class that deals heavily with three dimensions, and already I'm stumped. The inability to draw clear graphs by hand is crippling!

So here's the problem that's got me. It's really not Calculus, but I didn't know where to post this, and since the class I'm taking is Calc 3, I figured I might as well post it here... sorry if it's the wrong place.

So... if I understand correctly then EVERY point on the sphere must satisfy the requirement of having twice the distance from point A as from point B.
Since the whole three-dimensional thing is still so foreign to me, I figured I could get a better understanding by reducing it to 2 dimensions. So I took the cross-section of the plane containing the sphere's center and points A and B, and looked at the circle. I then applied a trial-and-error approach and made the circle centered on the origin with radius 1, then tried to come up with a configuration for points A and B to satisfy the original problem.
I wasn't able to do so.

It seems like the solution to the given condition would be something like an ellipsoid or hyperboloid... not a sphere. Am I missing something, or is this problem maybe just badly worded? Perhaps my prof meant "solid" instead of "sphere"?

Thanks!

Let $\displaystyle P=(x,y,z)$ be a point in our sphere, then $\displaystyle dist(P,A)=2\cdot dist(P,B)$ , so after squaring both sides we get:

$\displaystyle (x+2)^2+(y-5)^2+(z-2)^2=4[(x-5)^2+(y-2)^2+(z+3)^2)]$ , and opening parentheses and doing a little algebra we get:

$\displaystyle \left(x-\frac{22}{3}\right)^2+(y-1)^2+\left(z+\frac{14}{3}\right)^2=\frac{322}{9}$ , or something like this (check carefully this).

Tonio