Question

Find an equation of a sphere whose diameter has endpoints with coordinates (1,0,3) and (3,-5,6)

Answer 1

what's up dude

Answer 2

hey, what's this?

Answer 3

reviewing for my final

Answer 4

this was a question from the study guide i was never able to solve

Answer 5

find center point between them

(1,0,3) and (3,-5,6)

center of sphere
(2,-5/2,9/2)

Answer 6

(x-2)^2 + (y+5/2)^2+(z-9/2)^2=r^2

Answer 7

shouldn't i use the distance formula to get the radius?

Answer 8

for radius^2, (2-1)^2 + (-5/2)^2 +(9/2)^2

Answer 9

1+25/4 +81/4

Answer 10

(x-2)^2 + (y+5/2)^2+(z-9/2)^2=1+25/4 +81/4

Answer 11

so 96/4?

Answer 12

is r?

Answer 13

r^2

Answer 14

sphere formula

x^2 +y^2+z^2=r^2

Answer 15

so my equation would be in the form x^2 + y^2 + z^2 = (96/4)^2?

Answer 16

(x-a)^2 + (y-b)^2 + (z-a)^2 = (110/4)^2?

4+25+81=110

Answer 17

lol oops

Answer 18

alright...so how did you get ur center?

Answer 19

it looks like you subrtacted the x values....but then added y and z values and divided by 2

Answer 20

by finding center between two endpoints of diameter

Answer 21

midpoint; add points and divide by 2

Answer 22

ohhhhh lol

Answer 23

i see...didn't think that through clearly

Answer 24

(1,0,3) and (3,-5,6)

(1+3 )/2, (0-5)/2, (3+6)/2)

Answer 25

i go for the vertical format meself :)


(1, 0, 3)
(3,-5,6)
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(4,-5,9) / 2