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

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

Mathematics
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http://www.regentsprep.org/regents/math/algtrig/atc1/circlelesson.htm
sorry i gtg ^ that links explains what you have to do
awh man :c

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its ok i'll try to figure it out thanks tho :)
@pooja195 help me please :((
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 \((h,k)\)
start by looking at just the terms involving \(x\), i.e.:\[x^2+2x\]can you transform this into a form like \((x+?)^2\)?
i'll try, sorry im not good at this
im back
we know that:\[(x+a)^2=x^2+2ax+a^2\]
and we want to get:\[x^2+2x\]
well like the website says, you need to transform it into the center radius form by completing the square
can you see what value of \(a\) we need to pick?
nvr mind i will let @asnaseer explain :P
what do u mean @asnaseer
thanks though :)) @acxbox22
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?
is it 1?
correct! :)
phew than God i felt so stupid lol
so we know that:\[(x+1)^2=x^2+2x+1\]agreed so far?
yes :)
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?
i.e. we know:\[(x+1)^2=x^2+2x+1\]\[\therefore (x+1)^2-1=x^2+2x+1-1=x^2+2x\]
ohhh so now we're left with \[x^2+2x\] ?
exactly
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?
yes :)
good, now lets first simplify this by adding 1 to both sides to get rid of the -1 from the left-hand-side
\[(x+1)^2-1+y^2+4y=20\]\[\therefore (x+1)^2-1+y^2+4y+1=20+1\]\[(x+1)^2+y^2+4y=21\]following so far?
yes :)
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 \(4y\)?
2 :3
perfect! :)
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?
yes :D
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?
agreed c:
now again we can simplify this bu adding 4 to both sides as follows:\[(x+1)^2+(y+2)^2-4=21\]\[\therefore (x+1)^2+(y+2)^2-4+4=21+4\]\[\therefore (x+1)^2+(y+2)^2=25\]
agreed?
i get it now!! so the center will be (-1,-2) and the radius will be 5? :D
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
OMG THANK YOU SO MUCHHH!!!! you're a life saver, i really appreciate your help :))) <3
yw my friend. I really appreciate people who are genuinely interested in learning - keep up the good work! :)
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
ok, bye for now - take care my friend :)
thx and you too :)

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