Question

so for (x-3)^2+(y+1)^2=4
h=3,k=-1
r=2
right @satellite73

Answer 1

@Luigi0210

Answer 2

@johnweldon1993

Answer 3

yeah that looks right to me!
Center is (h,k) Radius (r)
h = 3
k = -1
r = 2

So yeah looks correct to me

Answer 4

ty!!!!! and what do you think about x^2+(y-2)^2=1?
r=
h=
k=

Answer 5

would h be 0?@johnweldon1993

Answer 6

Yeah h would be 0....k would be 2 (so the center would be at point (0,2) on a graph...

what about r...?

Answer 7

r=1

Answer 8

Perfect :)

Answer 9

U r the best thank you!

Answer 10

So basically something like that lol

Answer 11

And no problem! :)

Answer 12

yea looks right lol :)

Answer 13

just one more question, what if the problem looks like 2x^2+2y^2+20x+16y+74=0?

Answer 14

I hate those lol...hang on let me figure that out...

Answer 15

lol k

Answer 16

Lol it's a factoring process but I can't QUITE remember how to do it...still working on it though lol

Answer 17

ok thanks no worries i am trying to as well :)

Answer 18

is it 2(x+5)^2+2(y+4)^2-8=0 @johnweldon1993

Answer 19

-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 = ?

Answer 20

h=5
k=4
r=2

Answer 21

check the h and k again...

Answer 22

h=0
k=2
r=1

Answer 23

but I dont think I factored it correctly X(

Answer 24

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....

Answer 25

lol I check it with wolfram and got something else

Answer 26

Yeah wolfram has the same thing

h at -5
k at -4
and the radius looks to be 2

Answer 27

ok guess I gotta go with it

Answer 28

lmao really ?

http://www.wolframalpha.com/input/?i=2x%5E2%2B2y%5E2%2B20x%2B16y%2B74%3D0%3F

Answer 29

haha yup but thank you so much !!!!!!!!!!!!!!!!!!!!!!!@johnweldon1993