Questions: 3D Coordinate System - Spheres Part 1 of 2 Find an equation of a sphere if one of its diameters has endpoints (5,4,5) and (7,8,7). center: (x, y, z)=(6,6,6) radius: r=sqrt(6) Part 2 of 2 Write an equation of the sphere. (x-6)^2+(y-6)^2+(z-6)^2=6

Find the center and radius of the sphere given the endpoints of its diameter.

Find the center of the sphere.

The center of the sphere is the midpoint of the diameter with endpoints \((5, 4, 5)\) and \((7, 8, 7)\). The midpoint formula is:
\[ \text{Center} = \left( \frac{5 + 7}{2}, \frac{4 + 8}{2}, \frac{5 + 7}{2} \right) = (6, 6, 6).
\]

Find the radius of the sphere.

The radius is half the length of the diameter. First, calculate the distance between the endpoints using the distance formula:
\[ \text{Distance} = \sqrt{(7 - 5)^2 + (8 - 4)^2 + (7 - 5)^2} = \sqrt{4 + 16 + 4} = \sqrt{24}.
\]
The radius is half of this distance:
\[ r = \frac{\sqrt{24}}{2} = \sqrt{6}.
\]

The center of the sphere is \(\boxed{(6, 6, 6)}\), and the radius is \(\boxed{\sqrt{6}}\).

Write the equation of the sphere.

Write the equation using the center and radius.

The standard equation of a sphere with center \((h, k, l)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2.
\]
Substituting the center \((6, 6, 6)\) and radius \(\sqrt{6}\):
\[ (x - 6)^2 + (y - 6)^2 + (z - 6)^2 = 6.
\]

The equation of the sphere is \(\boxed{(x - 6)^2 + (y - 6)^2 + (z - 6)^2 = 6}\).

The center of the sphere is \(\boxed{(6, 6, 6)}\), and the radius is \(\boxed{\sqrt{6}}\).
The equation of the sphere is \(\boxed{(x - 6)^2 + (y - 6)^2 + (z - 6)^2 = 6}\).