anonymous
  • anonymous
so for (x-3)^2+(y+1)^2=4 h=3,k=-1 r=2 right @satellite73
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@Luigi0210
anonymous
  • anonymous
@johnweldon1993
johnweldon1993
  • johnweldon1993
yeah that looks right to me! Center is (h,k) Radius (r) h = 3 k = -1 r = 2 So yeah looks correct to me

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anonymous
  • anonymous
ty!!!!! and what do you think about x^2+(y-2)^2=1? r= h= k=
anonymous
  • anonymous
would h be 0?@johnweldon1993
johnweldon1993
  • johnweldon1993
Yeah h would be 0....k would be 2 (so the center would be at point (0,2) on a graph... what about r...?
anonymous
  • anonymous
r=1
johnweldon1993
  • johnweldon1993
Perfect :)
anonymous
  • anonymous
U r the best thank you!
johnweldon1993
  • johnweldon1993
|dw:1370978772149:dw| So basically something like that lol
johnweldon1993
  • johnweldon1993
And no problem! :)
anonymous
  • anonymous
yea looks right lol :)
anonymous
  • anonymous
just one more question, what if the problem looks like 2x^2+2y^2+20x+16y+74=0?
johnweldon1993
  • johnweldon1993
I hate those lol...hang on let me figure that out...
anonymous
  • anonymous
lol k
johnweldon1993
  • johnweldon1993
Lol it's a factoring process but I can't QUITE remember how to do it...still working on it though lol
anonymous
  • anonymous
ok thanks no worries i am trying to as well :)
anonymous
  • anonymous
is it 2(x+5)^2+2(y+4)^2-8=0 @johnweldon1993
johnweldon1993
  • johnweldon1993
-8 of course....forgot about that part XD yes that makes sense Good job! Then dividing everything by 2 makes it (x + 5)² + (y + 4)² = 4 So h = ? k = ? r = ?
anonymous
  • anonymous
h=5 k=4 r=2
johnweldon1993
  • johnweldon1993
check the h and k again...
anonymous
  • anonymous
h=0 k=2 r=1
anonymous
  • anonymous
but I dont think I factored it correctly X(
johnweldon1993
  • johnweldon1993
No....with the equation 2(x+5)^2+2(y+4)^2-8=0 *you factored it correctly, I have the same thing....but you should see that h = -5 k = -4 and r = 2 Let me double check with wolfram....
anonymous
  • anonymous
lol I check it with wolfram and got something else
johnweldon1993
  • johnweldon1993
Yeah wolfram has the same thing h at -5 k at -4 and the radius looks to be 2
anonymous
  • anonymous
ok guess I gotta go with it
johnweldon1993
  • johnweldon1993
lmao really ? http://www.wolframalpha.com/input/?i=2x%5E2%2B2y%5E2%2B20x%2B16y%2B74%3D0%3F
anonymous
  • anonymous
haha yup but thank you so much !!!!!!!!!!!!!!!!!!!!!!!@johnweldon1993

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