I am not sure how to proceed with this question from Stewart's SV Calculus text:
Find equations of the sphere's with center $(2, -3, 6)$ that touch (a) the $xy$-plane, (b) the $yz$-plane, (c) the $xz$-plane.
I know that the equation for a sphere should look like this: $$(x - 2)^2 + (y + 3)^2 + (z - 6)^2 = r^2$$
But I am not sure how to solve for $r^2$ to satisfy each of the above requirements. I noticed that the book's answers for $r^2$ are 36, 4, and 9 for $xy, yz$ and $xz$ (respectively). It appears that, to find the radius for each plane, I should just calculate the square of the axis that doesn't appear in the given plane. Is this really all there is to it? Can someone give me a more thorough explanation as to why this is the case?
If the circle touches the $xy$ plane, for example, then this means you can figure out - without any computation - the $(x,y,z)$ coordinates of the point where the circle touches the plane.
In particular, the $x,y$ coordinates must be the same as the center, and since you are on the $xy$ plane, the $z$ coordinate is $0$.
Then you just plug in the coordinates, and get:
$$(2 - 2)^2 + (-3 + 3)^2 + (0 - 6)^2 = r^2$$
From this you get $r=6$ (or $r^2=36$).
The other circles follow the same procedure, mutatis mutandis.