Question

How do you find an equation for each sphere that passes through the point (5, 1, 4) and is tangent to all three coordinate planes?

Answer 1

Use the standard Cartesian equation of a sphere:
`(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2" [1]"`

Explanation:

Consider the case where the sphere is tangent to the x-y plane. The plane touches the sphere with a radius that is perpendicular to the plane where the z coordinate is 0 at the point `(x_0,y_0,0)`. Substituting this point into equation [1]:

`(x_0-x_0)^2 + (y_0-y_0)^2 + (0-z_0)^2 = r^2`

`(-z_0)^2 = r^2`

Because only positive values of r are allowed, we have the equation.

`r = z_0`

Using symmetry the other points for the other two coordinate planes are, `(x_0,0,z_0), and (0,y_0,z_0)`.

This gives us the equation:

`r = x_0 = y_0 = z_0`

This allows us to reduce equation [1] to:

`(x-r)^2 + (y - r)^2 + (z-r)^2 = r^2" [2]"`

Substitute the point `(5,1,4)` into equation [2]:

`(5-r)^2 + (1 - r)^2 + (4-r)^2 = r^2`

Expand the squares:

`r^2 -10r + 25 + r^2 -2r + 1 + r^2 - 8r + 16 = r^2`

`2r^2 -20r + 42 = 0`

`r^2 - 10 + 21 = 0`

`(r -3)(r-7) =0`

`r = 3 and r = 7`

The equations are:

`(x - 3)^2 + (y - 3)^2 + (z-3)^2 = 3^2`

and

`(x - 7)^2 + (y - 7)^2 + (z-7)^2 = 7^2`