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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?

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    That's right. So, if the circle is to be tangent to the $x$-axis, then it must have a radius of $3$, i.e., $(x-2)^2+(y-3)^2=3^2$. A circle with the same centre but tangent to the $y$-axis would be $(x-2)^2+(y-3)^2=2^2$. Try drawing them! The analogous case in 3 dimensions should be easy once you see how this works in 2-d.2012-04-11

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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.

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The perpendicular distance from the center of the sphere to the xy plane will be 6 which will be equal to the radius of the sphere since xy plane is tangent to the sphere. Similarily, from yz and zx plane the perpendicular distance will be 2 and 3 respectively and the radius of the sphere will be 2 and 3 respectively.