Ok, I am working on circles in college algebra here. I am supposed to write this in standard form and determine both the center and radius if it exists.

x^2+y^2-2x+6y-15=0

Question

Ok, I am working on circles in college algebra here.  I am supposed to write this in standard form and determine both the center and radius if it exists.

x^2+y^2-2x+6y-15=0

anonymous · Answer

Rewrite:
x^2-2x+y^2+6y=15

Then complete the square:

x^2-2x+1+y^2+6y+9=15+1+9

(x-1)^2+(y+3)^2=25

anonymous · Answer

(x - 1)^2 + ( y + 3)^2 = 5^2

anonymous · Answer

do you know how to "complete the square?"

anonymous · Answer

( 1, -3) are centre 's coordinates ........ 5 is the radius

anonymous · Answer

both answer are correct.  do you know how to get them?

anonymous · Answer

@satellite... I thought I did but apparently I am not clear on it...

anonymous · Answer

lets rewrite as 
$$x^2-2x+y^2+6y=15$$ as step one

anonymous · Answer

step two is to think 
"half of -2 is -1, and -1 squared is 1"
half of 6 is 3 and 3 squared is 9

anonymous · Answer

yeah, I had it worked out that far...

anonymous · Answer

and then write 
$$(x-1)^2+(y+3)^2=15+1+9$$

anonymous · Answer

that is it.  so you have the form you want it in.

anonymous · Answer

oh geez.... I was soooo stinkin' close, thanks!

anonymous · Answer

yw