Question

Find the equation of the circle with center at (3, 2) and through the point (5, 4).

Answer 1

ponies!!! :D

Answer 2

hey! hot Rarity :D lol but i really need an answer

Answer 3

I only know part of it :(

Answer 4

I haven't touch this problem in eons x.x

Answer 5

x^2+y^2 = 1 is the default equation of the circle. the center is (0,0) and the radius is one.

Answer 6

since the center is at 3,2 it's (x-3)^2+(y-2)^2
but to find the radius.. x.x.

Answer 7

D: arghhh don't use the math you lose the math sad bambi@!

Answer 8

@Kainui

Answer 9

D: I can't help it if higher math courses only use some topics DKKKK

Answer 10

Elementary Linear Algebra had nothing to do with this D:

Answer 11

this shows up again in Calculus III, but I had the simple questions x.x

Answer 12

find radius
\[r^2= (x_b-x_a)^2+ (y_b-y_a)^2\]

Answer 13

to find the radius
\[\sqrt{(5-3)^2(4-2)^2}=r\]

Answer 14

so r = err add a plus

Answer 15

\[\sqrt{(5-3)^2+(4-2)^2}=r\]

Answer 16

i got 6

Answer 17

\[r=2\sqrt{2}\\so\\(x-3)^2+(2-y)^2=2\sqrt{2}\]

Answer 18

nope, in equation of a circle is r^2

Answer 19

so it schould be 8

Answer 20

instead of 2sqrt2

Answer 21

\[(x-3)^2+(y-2)^2=(2\sqrt{2})^2=8\]

Answer 22

sorry its late

Answer 23

DISTANCE FORMULA UGHHHHHHHHHH! *bangs head* of course! Use that and then square it... gawddd D((((

Answer 24

so then the equation of a circle that passes through those points is...?

Answer 25

\[(x-3)^2+(y-2)^2=8\]

Answer 26

yuppppp