Just a circle problem

  • #1
Hi, Math forums!
I need some help with a circle question.

3x^2 + 12x + 3y^2 - 5y = 2
And I was supposed to find the radius and center of the circle, So I first divided by 3:

x^2 +4x + y^2 - 5/3y = 2/3
And then I complete the square

x^2 + 4x + 4 + y^2 - 5/3y + 25/36 = 2/3 + 12/3 + 25/36 = 193/36

so then

(x+2)^2 + (y - 5/6)^2 = 193/36

and I don't know how I can get the standard (x-h)^2 - (y-k)^2 format from this.
Can anyone please help me? Thanks!
 

Answers and Replies

  • #2
jhae2.718
Gold Member
1,161
20
It's a circle, so the standard equation should be [tex](x-h)^2+(y-k)^2 = r^2[/tex].
 
  • #3
62
0
(x+2)^2 + (y - 5/6)^2 = 193/36

That is in the standard form for a circle, is it not? Just take the square root of 193/36 to get the radius...
 

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