# Find an equation of the sphere that passes through the origin and whose center is (5, 10, -9). Help with Calculus III

I have never seen a problem like this before, so I was wondering if anyone could give me help getting started. I'm studying for a quiz on Wednesday.

Find an equation of the sphere that passes through the origin and whose center is (5, 10, -9).

___ = 0 Note that you must put everything on the left hand side of the equation and that we desire the coefficients of the quadratic terms to be 1.

Similarly, how would one approach this:

Find an equation of the largest sphere with center (2, 10 , 4) that is contained completely in the first octant.

= 0 Note that you must move everything to the left hand side of the equation that we desire the coefficients of the quadratic terms to be 1.

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Though you say you are studying for Calculus III, the problem is really an analytic geometry problem, not a calculus problem. – Arturo Magidin Jan 25 '11 at 5:02

Let $r$ be the radius of the sphere. Then the equation of the sphere with center $(5,10,-9)$ is $$(x-5)^{2} + (y-10)^{2} + (z+9)^{2} =r^{2}$$

Now since your sphere passes through the origin$=(0,0,0)$, therefore you get $r^{2}=5^{2}+10^{2}+9^{2}=25+100+81=206$. Therefore the equation of the sphere is: $$(x-5)^{2}+(y-10)^{2}+(z-9)^{2}=206$$

For the next one:

If the sphere is in the 1st octant, the largest possible sphere must be tangent to either the $xy, xz$, $yz$ planes. That is, the radius of the sphere must be such that the surface of the sphere just touches one of the planes.

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Thank you very much for the advice, it worked great. However, I'm still stuck on the 1st octant problem -- I'm not sure how to determine the radius in that regard, I got sqrt(120) following the previous method. – Math Student Jan 25 '11 at 5:31
@Anonymous: As I hinted, you need to figure out which of the three planes (the $xy$ plane, the $yz$ plane, and the $xz$ plane) is closest to the center of the sphere. The distance to the closest plane is the radius. – Arturo Magidin Jan 25 '11 at 14:47

The radius of a sphere is the distance between the center and any point on the sphere. You know the center, and you know a point on the sphere, so the distance between the two points gives you the radius $r$.

Once you know the radius and the center, the equation of the sphere is very simple. It's almost like the case of a circle on the plane.

To jog your memory: the circle with center $(a,b)$ and radius $r$ has equation $$(x-a)^2 + (y-b)^2 = r^2.$$

Do you recall now what the equation of a sphere with radius $r$ and center $(a,b,c)$ is?

And since you know the center, and now you know how to find the radius, then the equation of the sphere is...

And as far as moving everything to one side, well, in the case of the circle I gave above, you can just as well write $(x-a)^2 + (y-b)^2 = r^2$ as $(x-a)^2 + (y-b)^2 - r^2 = 0$.

For the second problem, again let's think in terms of the plane. Suppose you want the equation of the circle with center at $(2,10)$ and completely contained in the first quadrant. How would we proceed? Imagine the largest circle: it must just touch at least one of the axes; how far is $(2,10)$ from the axes? It is $2$ away from the $y$-axis, and $10$ away from the $x$-axis. So the largest the radius can be is $2$, otherwise it would "cut through" the $y$-axis. So you'd want the circle with radius $2$ and center in $(2,10)$.

How would you go about doing this for the sphere in the first octant?

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