Chapter 9.6, Problem 68AYU

The Sphere In space, the collection of all points that arc the same distance from some fixed point is called a sphere. See the illustration. The constant distance is called the radius, and the fixed point is the center of the sphere. Show that an equation of a sphere with center at $\left(x_0,y_0,z_0\right)$ and radius $r$ is

${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(z-z_0\right)}^2=r^2$

[Hint: Use the Distance Formula (1).]

To determine

To find: Show that an equation of a sphere.

Answer to Problem 68AYU

Solution:

${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(z-z_0\right)}^2=r^2$

Explanation of Solution

Given:

${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(z-z_0\right)}^2=r^2$

Calculation:

The distance of any point on the sphere from the origin,

$\sqrt{{\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(z-z_0\right)}^2}$

This is the radius $r$ of the sphere.

Therefore, the equation of a sphere is:

${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(z-z_0\right)}^2=r^2$