# Thread: maximum distance from point to sphere

1. ## maximum distance from point to sphere

I have to calculate the maximum distance from the point $\displaystyle T(1, -1, 3)$ to the sphere $\displaystyle x^2+y^2+z^2-6x+4y-10z-62=0$.

Here's what I did.
The equation for the sphere is $\displaystyle (x-3)^2+(y+2)^2+(z-5)^2=100$.
That means the center of the sphere is $\displaystyle C(3, -2, 5)$, and the radius is $\displaystyle r=10$.
I think the maximum distance $\displaystyle d_m$ is the distance from the point $\displaystyle T$ to the center of the sphere plus the radius:
$\displaystyle d_m=d(T, C)+r$.

Is this okay?
Thank you!

2. Everything looks good so far. What do you get?

3. I get
$\displaystyle d((1, -1, 3), (3, -2, 5))=\sqrt((3-1)^2+(-2+1)^2+(5-2)^2 )=\sqrt(9)=3$ so $\displaystyle d_m=3+10=13$.

It seems too simple, are you sure I'm not thinking something terribly wrong?

4. You're not computing the distance formula correctly. It should be more like

$\displaystyle \sqrt{(3-1)^2+(-2+1)^2+(5-2)^2}=\sqrt{4+1+9}=\sqrt{14}.$

[EDIT]: Ignore this post. See below for correction.

5. Why do we have (5-2)^2 instead of (5-3)^2? I'm afraid I don't understand.

6. I'm sorry. You're correct. It should be 3, like you got.

7. Thank you so much! I'm quite insecure when I'm doing the exercises on my own, it really helps when someone checks your work.

8. If you feel insecure when you're doing exercises, I would come up with a system of checks you can run against your answers to justify them to yourself. It's good problem-solving strategy anyway.

You're very welcome. Have a good one!