# Finding Spheres Tangent to Three Coordinate Planes: A Calc 3 Problem

In summary, to find each sphere that passes through the point (5,1,4) and is tangent to all three coordinate planes, you need to set up the equation with the center point (s,s,s) and the radius s, using the fact that the distance between a point and a coordinate plane is just one of its coordinates. Then, use the condition that the sphere passes through the given point to solve for s and find the equation of the sphere.
The question asks to find each sphere that passes through the point (5,1,4) and are tangent to all three coordinate planes.
How would I set that up?
I have no problem finding the equation of a sphere when they give the center point and the radius, of course, but what do I need to know to find these equations?

OK, suppose your at the center point of the sphere. You look out and you should see each of the coordinate planes at equal distances because the sphere is tangent to each of them, so the vector from the center to the coordinate plane is perpendicular to the coordinate plane, meaning that the distance along this vector is the minimal distance and therefore "the" distance. Now, the distance between a point and a coordinate plane is just one of its coordinates. For example, the distance between a point and the x-y plane is just the z-coordinate. Now, since all of these distances are equal, the coordinates of the center must all be the same number. Call this number s, so the center is at (s,s,s). s must also be the radius of the sphere since it is the distance from the center to the surface. Now the the equation of the sphere is:
$$(x-s)^2+(y-s)^2+(z-s)^2=s^2$$
Use the last condition that it passes through the point (5,1,4) to solve for s and you're done.

you're awesome

Thanks again. You are awesome.

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