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]"(x−x0)2+(y−y0)2+(z−z0)2=r2 [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)(x0,y0,0). Substituting this point into equation [1]:

(x_0-x_0)^2 + (y_0-y_0)^2 + (0-z_0)^2 = r^2(x0−x0)2+(y0−y0)2+(0−z0)2=r2

(-z_0)^2 = r^2(−z0)2=r2

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

r = z_0r=z0

Using symmetry the other points for the other two coordinate planes are, (x_0,0,z_0), and (0,y_0,z_0)(x0,0,z0),and(0,y0,z0).

This gives us the equation:

r = x_0 = y_0 = z_0r=x0=y0=z0

This allows us to reduce equation [1] to:

(x-r)^2 + (y - r)^2 + (z-r)^2 = r^2" [2]"(x−r)2+(y−r)2+(z−r)2=r2 [2]

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

(5-r)^2 + (1 - r)^2 + (4-r)^2 = r^2(5−r)2+(1−r)2+(4−r)2=r2

Expand the squares:

r^2 -10r + 25 + r^2 -2r + 1 + r^2 - 8r + 16 = r^2r2−10r+25+r2−2r+1+r2−8r+16=r2

2r^2 -20r + 42 = 02r2−20r+42=0

r^2 - 10 + 21 = 0r2−10+21=0

(r -3)(r-7) =0(r−3)(r−7)=0

r = 3 and r = 7r=3andr=7

The equations are:

(x - 3)^2 + (y - 3)^2 + (z-3)^2 = 3^2(x−3)2+(y−3)2+(z−3)2=32

and

(x - 7)^2 + (y - 7)^2 + (z-7)^2 = 7^2(x−7)2+(y−7)2+(z−7)2=72

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