Question

Please help me? I am so lost!

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?

Answer 1

expand out
x^2 + y^2 -2yk +k^2 - 12 = 0

compare with the other one
2yk = 2y

-11 = k^2 -12

work out k

pssstt (try k = 1)

Answer 2

Another way is just to complete the square for y

x^2 + (y^2 -2y +1) -1 -11 =0

Answer 3

where did the -2yk +k^2 - 12 come from?

Answer 4

Then it simplifies down to

x^2 + (y-1)^2 = 12

Answer 5

I don't even know how you got the original one

Answer 6

All you are doing is completing the square for y so it simplifies.

Answer 7

i dont know how to complete a square...

Answer 8

Completing the square is a very useful tool the comes handy in many situations and is most definitely worth learning!!!!!!!!!!!!

Here is a link to a website that goes over the basics instead of me having to type it out

http://www.purplemath.com/modules/sqrquad.htm

Answer 9

I hope this helps :)

Answer 10

thanks