Algebra question?

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- anonymous

Algebra question?

- schrodinger

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- anonymous

Decide whether or not the equation has a circle as its graph. If it does, give the center a\[x^2+y^2-6x+2y=-9\]nd the radius. If it does not, describe the graph.

- anonymous

I know I need to rewrite into center radius form but i only get this far:
\[(x^2-6x)+(y^2+2y)=-9\]

- nincompoop

what is the general form for circle?

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## More answers

- anonymous

\[(x-h)^2+(y-k)^2=r^2\] ?

- nincompoop

does yours look like that?

- anonymous

no:(

- nincompoop

http://www.mathsisfun.com/algebra/circle-equations.html
there you have it

- anonymous

\[(x-6)^2+(y-2)^2=-9^2\]

- nincompoop

can you expand the left hand side?

- anonymous

wait, it'd be x+6 right? not x-6

- nincompoop

you complete the squares

- anonymous

\[(x+6)^2+(y-2)^2=-9^2\]

- nincompoop

look at the link I shared it shows you step by step using similar problem

- anonymous

ok hang on

- anonymous

I am closing to finding the answer will you tell me if its right?

- anonymous

Ok, I got (3,-1) and now cant figure out to how find the radius?

- anonymous

- dehelloo

hmm https://www.youtube.com/watch?v=qh5hNY83UA4
is an accurate video that follows up the process

- anonymous

am I completely wrong? :/ I have trouble with longer problems. can anyone solve this exact problem so I can see where I went wrong?

- anonymous

Can anyone just give me answer at least? Trying to finish my homework up and this is my last problem left.

- anonymous

- anonymous

AFK.

- Nnesha

he died ?? ;-; nin!!

- anonymous

Haha I don't know what happened!

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @rawrolivia
\[(x+6)^2+(y-2)^2=-9^2\]
\(\color{blue}{\text{End of Quote}}\)
you didn't complete the square

- anonymous

I'll show you what I did. Give me a moment to type it all

- Nnesha

okay

- anonymous

\[x^2+y^2-6x+2y=-9 -->
(x^2-6x+9)+(y^2+2y+1)=-9+9+1\]
\[(x-3)^2+(y+1)^2=1\]
(3,-1)

- peridot

...I think I got the answer but I'm unsure. Sorry I've been meaning to reply for awhile but uh.. I probably got it wrong but this is what I got from it:
Center of the circle is: C = (-6, 2)
Radius of the circle is 9.

- peridot

- Nnesha

yes your work is correct
remember to complete the square we should we should divide b by 2
and then add (b/2)^2 to the opposite side
that's how u got +9 and 1 at right side
-9+9+1= 1 so yes your work is correct good job!

- anonymous

ok but @peridot got a different answer :/

- Nnesha

\[x^2+y^2-6x+2y=-9 -->
(x^2-6x+\color{Red}{9})+(y^2+2y+1)=-9+9+1\]
how did you get 9 ? explain

- anonymous

-6 = -3, -3^2 =9

- Nnesha

-6=-3 ? is there a typo ?

- anonymous

I thought you had to take half of the middle number, and then square it

- Nnesha

oh yes so -6/2 = -3
-6 isn't equal to -3 alright and yes (-3)^2 =9
and like i said you should (b/2)^2 to the right side \[x^2+y^2-6x+2y=-9 -->
(x^2-6x+\color{Red}{9})+(y^2+2y+1)=-9+\color{ReD}{9}+1\]

- Nnesha

and same thing with one you take half of the middle term 2/2 =1 and then take square add to the right side \[x^2+y^2-6x+2y=-9 -->
(x^2-6x+\color{Red}{9})+(y^2+2y+1)=-9+\color{ReD}{9}+\color{blue}{1}\]
-9+9 +1 = 1
so if your work is correct then obviously your answer is correct

- anonymous

ok but I need to find out what the radius is now

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @rawrolivia
\[(x-h)^2+(y-k)^2=r^2\] ?
\(\color{blue}{\text{End of Quote}}\)
general form of the circle write where r is the radius

- Nnesha

now compare your equation with that general form what is the radius in that equation ?

- anonymous

So radius is 1?

- anonymous

going to walmart, can i just please have answer? this is confusing me more than it should :(

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @rawrolivia
So radius is 1?
\(\color{blue}{\text{End of Quote}}\)
yes right

- Nnesha

you already know the answer..

- Nnesha

\[(x-h)^2+(y-k)^2=r^2\]
general form of the circle write where r is the radius and (h,k) is the center

- nincompoop

I went to school to attend a session.
Sorry. Let me know if you want to further learn or go over about circles.

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