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?

1 Answer
Douglas K.
Jul 30, 2017

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#

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