Question

HELPPPPPP ASAP PLEASE!! D:


Using the following equation, find the center and radius of the circle.
x2 + 2x + y2 + 4y = 20

Answer 1

http://www.regentsprep.org/regents/math/algtrig/atc1/circlelesson.htm

Answer 2

sorry i gtg ^ that links explains what you have to do

Answer 3

awh man :c

Answer 4

its ok i'll try to figure it out thanks tho :)

Answer 5

@pooja195 help me please :((

Answer 6

you need to transform this equation:\[x^2 + 2x + y^2 + 4y = 20\]into the standard form:\[(x-h)^2+(y-k)^2=r^2\]which gives a circle of radius \(r\) and center at \((h,k)\)

Answer 7

start by looking at just the terms involving \(x\), i.e.:\[x^2+2x\]can you transform this into a form like \((x+?)^2\)?

Answer 8

i'll try, sorry im not good at this

Answer 9

im back

Answer 10

we know that:\[(x+a)^2=x^2+2ax+a^2\]

Answer 11

and we want to get:\[x^2+2x\]

Answer 12

well like the website says, you need to transform it into the center radius form by completing the square

Answer 13

can you see what value of \(a\) we need to pick?

Answer 14

nvr mind i will let @asnaseer explain :P

Answer 15

what do u mean @asnaseer

Answer 16

thanks though :)) @acxbox22

Answer 17

we know that:\[(x+a)^2=x^2+2ax+a^2\]can you see what value of \(a\) we need to pick so that we get a \(2x\) in the expansion on the right-hand-side?

Answer 18

is it 1?

Answer 19

correct! :)

Answer 20

phew than God i felt so stupid lol

Answer 21

so we know that:\[(x+1)^2=x^2+2x+1\]agreed so far?

Answer 22

yes :)

Answer 23

but we need just \(x^2+2x\), so we need to subtract 1 from both sides to get:\[(x+1)^2-1=x^2+2x\]agreed?

Answer 24

i.e. we know:\[(x+1)^2=x^2+2x+1\]\[\therefore (x+1)^2-1=x^2+2x+1-1=x^2+2x\]

Answer 25

ohhh so now we're left with \[x^2+2x\] ?

Answer 26

exactly

Answer 27

now remember your original equation was:\[x^2 + 2x + y^2 + 4y = 20\]we can now substitute \(x^2+2x\) with the new expression we found to get:\[(x+1)^2-1+y^2+4y=20\]understand?

Answer 28

yes :)

Answer 29

good, now lets first simplify this by adding 1 to both sides to get rid of the -1 from the left-hand-side

Answer 30

\[(x+1)^2-1+y^2+4y=20\]\[\therefore (x+1)^2-1+y^2+4y+1=20+1\]\[(x+1)^2+y^2+4y=21\]following so far?

Answer 31

yes :)

Answer 32

great!

so now we have:\[(x+1)^2+y^2+4y=21\]lets now concentrate on the terms involving \(y\), we have:\[y^2+4y\]here again we know that:\[(y+b)^2=y^2+2by+b^2\]so can you think of what value we need for \(b\) so that the \(2by\) term becomes \(4y\)?

Answer 33

2 :3

Answer 34

perfect! :)

Answer 35

so we now have:\[(y+2)^2=y^2+4y+4\]but we are after just \(y^2+4y\), so we need to subtract 4 from both sides, i.e.:\[(y+2)^2=y^2+4y+4\]\[\therefore (y+2)^2-4=y^2+4y+4-4=y^2+4y\]agreed?

Answer 36

yes :D

Answer 37

ok, now recall we originally were given:\[x^2 + 2x + y^2 + 4y = 20\]which we transformed using \((x+1)^2-1=x^2+2x\) into:\[(x+1)^2+y^2+4y=21\]and now we also know that:\[(y+2)^2-4=y^2+4y\]so if we now replace \(y^2+4y\) with the new expression we just found for it, then we get:\[(x+1)^2+(y+2)^2-4=21\]agreed?

Answer 38

agreed c:

Answer 39

now again we can simplify this bu adding 4 to both sides as follows:\[(x+1)^2+(y+2)^2-4=21\]\[\therefore (x+1)^2+(y+2)^2-4+4=21+4\]\[\therefore (x+1)^2+(y+2)^2=25\]

Answer 40

agreed?

Answer 41

i get it now!! so the center will be (-1,-2) and the radius will be 5? :D

Answer 42

perfect!

I was going to add one more step saying you have to be careful now as we are trying to get to this form:\[(x-h)^2+(y-k)^2=r^2\]so we must rewrite our equation as:\[(x-(-1))^2+(y-(-2))^2=5^2\]but I guess you beat me to it! :D

Answer 43

OMG THANK YOU SO MUCHHH!!!! you're a life saver, i really appreciate your help :))) <3

Answer 44

yw my friend.

I really appreciate people who are genuinely interested in learning - keep up the good work! :)

Answer 45

ahh you're awesome!! I spend more than an hour trying to figure this out lol u solved it in less that half an hour XD

Answer 46

ok, bye for now - take care my friend :)

Answer 47

thx and you too :)