Question

Find the standard equation of the sphere with the given characteristics. endpoints of a diameter: (6, 1, 5), (1, 5, −1)

Answer 1

The first thing we must do for this case, is to find the center of the sphere, for this, we use the following equation:
 $C = (\frac{x1+x2}{2},\frac{y1 + y2}{2},\frac{z1+z2}{2})$
 Substituting values we have:
 $C = (\frac{6+1}{2},\frac{1 + 5}{2},\frac{5-1}{2})$
 Rewriting we have:
 $C = (\frac{7}{2},\frac{6}{2},\frac{4}{2})$
 $C = (3.5,3,2)$
 We search now, the radius of the sphere.
 For this, we use the formula of distance between points.
 We have then:
 $r = \sqrt{(x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2}$
 Substituting values we have:
 $r = \sqrt{(1-3.5)^2 + (5-3)^2 + (-1-2)^2}$
 $r = 4.4$
 Then, the standard equation is given by:
 $(x-xo)^2+(y-yo)^2+(z-zo)^2=r^2$
 Substituting values:
 $(x-3.5)^2+(y-3)^2+(z-2)^2=4.4^2$
 Rewriting we have:
 $(x-3.5)^2+(y-3)^2+(z-2)^2=19.36$
 Answer:
 the standard equation of the sphere is:
 $(x-3.5)^2+(y-3)^2+(z-2)^2=19.36$