### Video Transcript

Give the equation of the sphere of
center negative six, 15, 11 that touches the 𝑥𝑦- plane.

A sphere is a three-dimensional
shape where every point is a distance 𝑟, the radius of the sphere, from the
center. The equation of a sphere in
standard form is 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus
𝑐 all squared is equal to 𝑟 squared, where the radius of the sphere is 𝑟 and its
center has coordinates 𝑎, 𝑏, 𝑐. In this question, we are told that
the center has coordinates negative six, 15, 11. Therefore, these are the values of
𝑎, 𝑏, and 𝑐. We are also told that the sphere
touches the 𝑥𝑦-plane. The distance from the center of a
sphere to the 𝑥𝑦-plane is the absolute value or modulus of 𝑐. This means that the radius of our
sphere is the absolute value of 11, which is equal to 11.

Substituting in the values of 𝑎,
𝑏, 𝑐, and 𝑟, we have 𝑥 minus negative six squared plus 𝑦 minus 15 squared plus
𝑧 minus 11 squared is equal to 11 squared. Subtracting negative six is the
same as adding six, so the first term becomes 𝑥 plus six all squared. As 11 squared is equal to 121, the
equation of the sphere of center negative six, 15, 11 that touches the 𝑥𝑦-plane is
𝑥 plus six squared plus 𝑦 minus 15 squared plus 𝑧 minus 11 squared is equal to
121.

An alternative method to calculate
the radius of the sphere would be to recognize that when the sphere touches the
𝑥𝑦-plane, the 𝑧-coordinate must be equal to zero. The point that is closest to the
center of the sphere on the 𝑥𝑦-plane has coordinates negative six, 15, zero. We could then find the distance
between the center and this point, which once again would give us an answer of
11. This confirms that the radius of
the sphere is 11.