SOLUTION: Help?! I was searching throughout the internet for a video with a detailed explanation of Algebra Conics, but no luck D: Can anyone please help me with these problems? It's prac

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Question 616003: Help?!
I was searching throughout the internet for a video with a detailed explanation of Algebra Conics, but no luck D:
Can anyone please help me with these problems?
It's practice for my final, and I'm just panicking!!
Write the equation in standard form, find the center and radius. ((a CIRCLE))
1)x^2+y^2-12x+14y+17=0
ANY HELP IS SINCERELY AND GRATEFULLY APPRECIATED!
Found 2 solutions by richwmiller, Theo:
Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website!
(x-6)^2+(y+7)^2=68
circle
center | (6, -7)
radius | 2 sqrt(17)~~8.24621
diameter | 4 sqrt(17)~~16.4924
area | 68 pi~~213.628
perimeter | 4 sqrt(17) pi~~51.8125

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
your equation is x^2 + y^2 - 12x + 14y + 17 = 0
put the x's and the y's together to get:
x^2 - 12x + y^2 + 14y + 17 = 0
subtract 17 from both sides of this equation to get:
x^2 - 12x + y^2 + 14y = -17
group your x's and y's together as shown below:
(x^2 - 12x) + (y^2 + 14y) = -17
factor each group by completing the squares on them to get:
(x-6)^2 + (y+7)^2 = -17 + 36 + 49
combine like terms to get:
(x-6)^2 + (y+7)^2 = 68
the center is (6,-7)
the radius is sqrt(68)
here's some references for you.
http://www.regentsprep.org/Regents/math/algtrig/ATC1/circlelesson.htm
http://www.purplemath.com/modules/solvquad3.htm