Question

what is the standard form of the equation of the circle x^2-2x+y^2-8x+1=0

Answer 1

\[x^2 -2x+y^2-8x+1=0\] compelte the square for both x and y

Answer 2

err 8y?

Answer 3

its a race to see who can complete the square the fastest

Answer 4

\[x^2\color{red}{-2x}+y^2\color{red}{-8x}+1=0\]Combine like terms first.

Answer 5

Ahahaha.

Answer 6

What is (-2-8)x?

Answer 7

-10x right?

Answer 8

Yes.

Answer 9

\[x^2\color{red}{-2x}+y^2\color{red}{-8x}+1=0\]\[(x^2 \color{red}{-10x})+y^2 +1=0\]Now you want to complete the square inside the parenthesis by finding your new \(c\) value.

Answer 10

Basically you want to complete a quadratic inside the (), \(ax^2+bx+c\)

Answer 11

so finding \(c\) is easy, \(c=\left(\frac{b}{2}\right)^2 = \left(\frac{-10}{2}\right)^2 = (-5)^2 = 25\)

Answer 12

\[\color{blue}{c}=\left(\frac{b}{2}\right)^2 = \left(\frac{-10}{2}\right)^2 = (-5)^2 = \color{blue}{25}\]Now add that in .\[(x^2 \color{red}{-10x}+\color{blue}{25} )+y^2 +1-25=0\] Remember that when you add anything in to complete a square, you must also subtract by the same number outside the square.

Answer 13

Simplify it. \[(x-5)^2+y^2 -24=0\]Put the 24 onto the other side. \[(x-5)^2 +y^2 =-24\]

Answer 14

Standard form of a circle is: \((x-h)^2 +(y-k)^2 =r^2\)

Answer 15

Are we sure that should not have been 8y not 8x?

Answer 16

seems weird to test them on combining like variables at this stage...

Answer 17

We could do that too,I was just working with what was given, lol

Answer 18

I know. ;) was more for the user to make sure, then you can see another way!

Answer 19

wouldn't it be positive 24 not negative?

Answer 20

Oh yes you are right, \(1-25 = 0 \implies -24\) and on the other side of the 0 it becomes \(+24\)

Answer 21

Sorry.

Answer 22

Naw your good it was very helpful!!! thank you!

Answer 23

i gotta go, when I come back I'll show you how to work the equation the way @zzr0ck3r mentioned. Or he can show you as well!

Answer 24

yeah i still think that -8x is a typo

Answer 25

It's not... that is what is write on my paper.

Answer 26

To avoid simplifying your functon restricted only to x variables, your equation should have looked like: \( x^2-2x+y^2-8y+1=0\)

You would have completed the square then for BOTH x and y.\[(x^2-2x)+(y^2-8y)=0\]

Answer 27

You should definitely try it this way, it'll give you practice in finding your center, as well as competing squares for 2 variables instead of just 1.

Answer 28

@zzr0ck3r it's your turn. Take care you guys!

Answer 29

lol, if it was not 8y it don't matter....