Question

The circle given by x^2+y^2-2y-11=0 can be written in standard form like this: x^2+(y-k)^2=12. What is the value of k in this equation?

I'm not really sure what the question is asking me...

Answer 1

this is the cirlcle's formula/equation .. (x-h)+(y-k)^2=r^2

Answer 2

So, I'm not sure if I explain what k stands for or what I'm supposed to plug in for k... Assuming that it means what II plug in because of "What is the value of k in this equation?"
Would my k value be -2?

Answer 3

let me try to solve it

Answer 4

Alright. No problem. n_n
Thank yhu for helping me~

Answer 5

what is the subject title?

Answer 6

Pre-Calculus

Answer 7

x^2+y^2-2y-11=0
x^2+y^2-2y=-11
x^2+y^2-2y-1=-11
x^2+y^2-2y-1=-11-1
x^2+(y-1)^2=-12
from this equation(x-h)=(y-k)=r^2
the value of y would be -1
.. I dont know if you're using the same approach with mine.

Answer 8

Haha, I guess I wasn't. >~<
Thank yhu so much for the help. And taking the time to work that out

Answer 9

haha.. Im using analytic geometry for this one.

Answer 10

That's actually how my teacher worked it out. I can see where I went wrong by looking at what yhu did. I missed subtracting the 1

Answer 11

Adding the 1*
My bad.

Answer 12

so,what value did you got?

Answer 13

I got positive 2 to start with, I forgot to add the 1
When I did it again I got -1

Answer 14

Yhu had an error tho.

\[x^{2}+y^{2}-2y-11=0\]\[x^{2}+y^{2}-2y=11\]\[x^{2}+y^{2}-2y+1=11+1\]\[x^{2}+(y-1)^{2}=12\]

Answer 15

what part?

Answer 16

Yhu want to add 11 to each side, which will give yhu positive 11 on the other side of the = sign.