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