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Godlovesme

  • one year ago

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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  1. acxbox22
    • one year ago
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    http://www.regentsprep.org/regents/math/algtrig/atc1/circlelesson.htm

  2. acxbox22
    • one year ago
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    sorry i gtg ^ that links explains what you have to do

  3. Godlovesme
    • one year ago
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    awh man :c

  4. Godlovesme
    • one year ago
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    its ok i'll try to figure it out thanks tho :)

  5. Godlovesme
    • one year ago
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    @pooja195 help me please :((

  6. asnaseer
    • one year ago
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    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 \((h,k)\)

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

  8. Godlovesme
    • one year ago
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    i'll try, sorry im not good at this

  9. acxbox22
    • one year ago
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    im back

  10. asnaseer
    • one year ago
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    we know that:\[(x+a)^2=x^2+2ax+a^2\]

  11. asnaseer
    • one year ago
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    and we want to get:\[x^2+2x\]

  12. acxbox22
    • one year ago
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    well like the website says, you need to transform it into the center radius form by completing the square

  13. asnaseer
    • one year ago
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    can you see what value of \(a\) we need to pick?

  14. acxbox22
    • one year ago
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    nvr mind i will let @asnaseer explain :P

  15. Godlovesme
    • one year ago
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    what do u mean @asnaseer

  16. Godlovesme
    • one year ago
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    thanks though :)) @acxbox22

  17. asnaseer
    • one year ago
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    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?

  18. Godlovesme
    • one year ago
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    is it 1?

  19. asnaseer
    • one year ago
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    correct! :)

  20. Godlovesme
    • one year ago
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    phew than God i felt so stupid lol

  21. asnaseer
    • one year ago
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    so we know that:\[(x+1)^2=x^2+2x+1\]agreed so far?

  22. Godlovesme
    • one year ago
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    yes :)

  23. asnaseer
    • one year ago
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    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?

  24. asnaseer
    • one year ago
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    i.e. we know:\[(x+1)^2=x^2+2x+1\]\[\therefore (x+1)^2-1=x^2+2x+1-1=x^2+2x\]

  25. Godlovesme
    • one year ago
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    ohhh so now we're left with \[x^2+2x\] ?

  26. asnaseer
    • one year ago
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    exactly

  27. asnaseer
    • one year ago
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    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?

  28. Godlovesme
    • one year ago
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    yes :)

  29. asnaseer
    • one year ago
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    good, now lets first simplify this by adding 1 to both sides to get rid of the -1 from the left-hand-side

  30. asnaseer
    • one year ago
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    \[(x+1)^2-1+y^2+4y=20\]\[\therefore (x+1)^2-1+y^2+4y+1=20+1\]\[(x+1)^2+y^2+4y=21\]following so far?

  31. Godlovesme
    • one year ago
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    yes :)

  32. asnaseer
    • one year ago
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    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 \(4y\)?

  33. Godlovesme
    • one year ago
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    2 :3

  34. asnaseer
    • one year ago
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    perfect! :)

  35. asnaseer
    • one year ago
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    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?

  36. Godlovesme
    • one year ago
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    yes :D

  37. asnaseer
    • one year ago
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    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?

  38. Godlovesme
    • one year ago
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    agreed c:

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

  40. asnaseer
    • one year ago
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    agreed?

  41. Godlovesme
    • one year ago
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    i get it now!! so the center will be (-1,-2) and the radius will be 5? :D

  42. asnaseer
    • one year ago
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    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

  43. Godlovesme
    • one year ago
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    OMG THANK YOU SO MUCHHH!!!! you're a life saver, i really appreciate your help :))) <3

  44. asnaseer
    • one year ago
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    yw my friend. I really appreciate people who are genuinely interested in learning - keep up the good work! :)

  45. Godlovesme
    • one year ago
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    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

  46. asnaseer
    • one year ago
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    ok, bye for now - take care my friend :)

  47. Godlovesme
    • one year ago
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    thx and you too :)

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