Question

Find the equation of the sphere if one of its diameters has endpoints (8,7,6) and (9,9,9).

Answer 1

we are given

diameters has endpoints (8,7,6) and (9,9,9)

Center:

we will find mid point of these two points

`(a,b,c)=((8+9)/(2),(7+9)/(2),(6+9)/(2))`

`(a,b,c)=(8.5,8,7.5)`

Radius:

radius is half of distance between these two points

`d=√((9-8)^2+(9-7)^2+(9-6)^2)`

`d=√(14)`

now, we can find radius

`r=(d)/(2)`

`r=(√(14))/(2)`

Equation of sphere:

we can use formula

`(x-a)^2+(y-b)^2+(z-c)^2=r^2`

now, we can plug values

and we get

`(x-8.5)^2+(y-8)^2+(z-7.5)^2=((√(14))/(2))^2`

`(x-8.5)^2+(y-8)^2+(z-7.5)^2=(14)/(4)`.............Answer

Answer 2

we know that

the equation of the sphere is equal to

`(x-h)^(2) +(y-k)^(2)+(z-l)^(2)=r^(2)`

where

(h,k,l) is the center of the sphere

r is the radius of the sphere

Step 1

Find the center of the sphere (h,k,l)

Let

A(8,7,6) B(9,9,9)

we know that

the midpoint of the diameter is the center point

Find the midpoint AB

in the x-coordinate

=(8+9)/2------> 8.5

in the y-coordinate

=(7+9)/2------> 8

in the z-coordinate

=(6+9)/2------> 7.5

the center is (8.5,8,7.5)

Step 2

Find the length of the diameter (distance AB)

A(8,7,6) B(9,9,9)

`d=\sqrt{(x-x1)^(2)+(y-y1)^(2)+(z-z1)^(2)}`

`dAB=\sqrt{(9-8)^(2)+(9-7)^(2)+(9-6)^(2)}`

`dAB=\sqrt{(1)^(2)+(2)^(2)+(3)^(2)}`

`dAB=√(14)\ units`

the diameter is
`√(14)\ units`

the radius is
`r=(√(14) )/2\ units`

Step 3

Find the equation of the sphere

`(x-h)^(2) +(y-k)^(2)+(z-l)^(2)=r^(2)`

`r=(√(14) )/2\ units`

the center is (8.5,8,7.5)

Substitute

`(x-8.5)^(2) +(y-8)^(2)+(z-7.5)^(2)=(√(14) )/2)^(2)`

`(x-8.5)^(2) +(y-8)^(2)+(z-7.5)^(2)=(14/4)`

`(x-8.5)^(2) +(y-8)^(2)+(z-7.5)^(2)=(3.5)`

therefore

the answer is

The equation of the sphere is

`(x-8.5)^(2) +(y-8)^(2)+(z-7.5)^(2)=(3.5)`