Question

Find the center of a circle with equation x^2+y^2-6x+4y-4=0

Answer 1

Use the "completing the square" method
group similar terms and transpose all constants to the right side of the equation first...

Answer 2

what are constants

Answer 3

A.) (3,2)
B.) (3,-2)
C.) (-3,2)
D.) (-3,-2)

Answer 4

Those without variables( 2, 3, 4 )...are you in a hurry? do you need the answer right away?

Answer 5

im taking my graduation test but i have time

Answer 6

im not very good at learning something over the computer

Answer 7

Oh, i see..so ,what is the constant in the equation? x^2+y^2-6x+4y-4=0

Answer 8

um 6x and 4y ?? can there only be one?

Answer 9

or was the constant the number that didnt have a letter in it?? i cant remember

Answer 10

no....constant are only composed of numbers (they don't have x's or y's) well at the state of our equation there is only one but there will be later when we expand it

Answer 11

i mean yeah, those without letters are constants

Answer 12

ok so its -4

Answer 13

yes , now when you transpose it to the other side, its sign changes

Answer 14

ok so its a positive 4 now. do i just move the constants to the right side?

Answer 15

cause its already on the right side

Answer 16

yup can you show me the equation?

Answer 17

um.. X^2+y^2-6x+4y+4=0

Answer 18

with the constants on the right side..

Answer 19

it is on the right side

Answer 20

oh what i meant was at the right side of the = sign (that's the right side of the equation)

Answer 21

so it would be =0+4 on the right side? im sorry im really bad at this

Answer 22

yeah but you don't need to write the 0 since anything added to 0 is equal to itself...now what is our equation so far?

Answer 23

oh ok
x^2+y^2-6x+4y=4

Answer 24

correct :D the next thing to do is group the x and y's together inside parentheses
( the one's with x go here ) + ( the ones with y here ) =4

Answer 25

(x^2-6x) + (y^2+4y)=4

Answer 26

ok we're almost done . we need to complete the square of the terms inside the parentheses, to do that that divide the second term by two and square it, then add it inside the parentheses for each. also add that to the right side

example
(x^2-2x+ 1) - 1 because -2/2 =-1 and -1^2 =1
(y^2-8x+16) 16 because -8/2 =-4 and 4^2= 16

(x^2-2x+1) +(y^2-8x+16)= 3 +1+16

Answer 27

3+2=4 ?

Answer 28

So, (3,2) ?

Answer 29

Well its close but here's how you do it
(x^2-6x+9) +(y^2+4y+4) = 4+9+4
^
(x-3)^2 + (y+2)^2= 17
So the answer is 3,-2

Answer 30

oh ok thanks :)