HELPPPPPP ASAP PLEASE!! D:
Using the following equation, find the center and radius of the circle.
x2 + 2x + y2 + 4y = 20

- Godlovesme

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- acxbox22

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

- acxbox22

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

- Godlovesme

awh man :c

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## More answers

- Godlovesme

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

- Godlovesme

@pooja195 help me please :((

- asnaseer

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 \((h,k)\)

- asnaseer

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

- Godlovesme

i'll try, sorry im not good at this

- acxbox22

im back

- asnaseer

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

- asnaseer

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

- acxbox22

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

- asnaseer

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

- acxbox22

nvr mind i will let @asnaseer explain :P

- Godlovesme

what do u mean @asnaseer

- Godlovesme

thanks though :)) @acxbox22

- asnaseer

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?

- Godlovesme

is it 1?

- asnaseer

correct! :)

- Godlovesme

phew than God i felt so stupid lol

- asnaseer

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

- Godlovesme

yes :)

- asnaseer

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?

- asnaseer

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

- Godlovesme

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

- asnaseer

exactly

- asnaseer

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?

- Godlovesme

yes :)

- asnaseer

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

- asnaseer

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

- Godlovesme

yes :)

- asnaseer

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 \(4y\)?

- Godlovesme

2 :3

- asnaseer

perfect! :)

- asnaseer

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?

- Godlovesme

yes :D

- asnaseer

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?

- Godlovesme

agreed c:

- asnaseer

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

- asnaseer

agreed?

- Godlovesme

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

- asnaseer

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

- Godlovesme

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

- asnaseer

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

- Godlovesme

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

- asnaseer

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

- Godlovesme

thx and you too :)

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