Question

will fan and medal

Answer 1

@The_Jokers_wife

Answer 2

what?

Answer 3

how are u?

Answer 4

okay u?

Answer 5

i summoned the courage to actually ask you. I am feeling great.

Answer 6

sure

Answer 7

@KonaFirestar @The_Jokers_wife Please do not spam someone else's question. Use the chat instead.

Answer 8

@cassandralombardi Hold on a sec while i read your question.

Answer 9

XD @StudyGurl14 don't u see we stop posting things in the chat XD u should have said that 2 min's ago #reading problem

Answer 10

??? @The_Jokers_wife

Answer 11

@cassandralombardi
you have \[x^2 + y^2 -6y -12 = 0\]
Now, put
\[y^2 -6y-12 = (y-k)^2\]
form

Answer 12

@cassandralombardi
You're going to need to complete the square. Do you know how to do that?

Answer 13

To complete the square, find this value, then add it to both sides of the equation.
\(\large (\frac{-6}{2})^2\)

Answer 14

By adding that value, you can easier factor the quadratic

Answer 15

right

Answer 16

so you have
\(\large y^2 -6y-3\)
Wait...something not right...that's not factor-able

Answer 17

@sangya21 Can you do this? I'm confused...

Answer 18

\[x^2 = -y^2 +6y +12\]
\[x^2 = -(y -k)^2 +21 \]
\[-y^2 +6y +12 = -(y -k)^2 +21 \]
\[-y^2 +6y +12 - 21 = -(y -k)^2 \]
\[-y^2 +6y -9 = -(y -k)^2 \]
\[y^2 -6y + 9 = (y -k)^2 \]
\[y^2 -3y -3y +9 = (y -k)^2 \]
\[y(y -3) -3(y -3) = (y -k)^2 \]
\[(y -3)(y -3) = (y -k)^2 \]
\[(y -3)^2 = (y -k)^2 \]

Answer 19

Is this correct? @StudyGurl14

Answer 20

yeah

Answer 21

sure.

Answer 22

omg @sangya21 that's amazing! EXCELLENT job!

Answer 23

Well sometimes confusion helps. @StudyGurl14 you showed the way. ;)

Answer 24

standard form of equation
(a,b) are the center. r = radius.
\[(x-a)^2 +(y-b)^2 = r^2\]

Answer 25

lol, thx. That's kind of you. :)

Answer 26

@cassandralombardi have you done it?

Answer 27

Look at the graph which point is at the center of the circle