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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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Note that the question is looking for three different spheres, not one sphere that satisfies all three constraints. It might help to consider an analogy in 2 dimensions: what would be the equation of a circle with centre $(2,3)$ that was tangent to the $x$-axis? –  Théophile Apr 10 '12 at 21:47
    
Right, I know that I am finding an equation for three different spheres. In the case of 2 dimensions, it would look similar to the above equation without the z-axis, i.e. $(x - 2)^2 + (y - 3)^2 = r^2$. Correct? –  Dylan Apr 10 '12 at 22:51
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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. –  Théophile Apr 11 '12 at 1:59

1 Answer 1

up vote 2 down vote accepted

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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