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artofspeed

  • 3 years ago

Find the center and radius of this equation of circle x^2+y^2-ax+by=0

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  1. AccessDenied
    • 3 years ago
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    Well, this doesn't look like much of a nice circle equation. I prefer \((x - h)^2 + (y - k)^2 = r^2\), which has center (h, k), and radius r. I think if we can write it like that, we'll have more progress! Are you familiar with completing the square in quadratics?

  2. artofspeed
    • 3 years ago
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    i got (x-a)^2 + (y+b)^2 = (a/2)^2 + (b/2)^2

  3. artofspeed
    • 3 years ago
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    so the radius must be \[\sqrt{(a/2)^2 + (b/2)^2}\] and center is (a, -b) right?

  4. AccessDenied
    • 3 years ago
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    I think I agree with your radius, but if we expand out (x - a)^2, we'd get x^2 - 2a x + a^2. Instead, I think (x - a/2)^2 would be more accurate. It expands to x^2 - a/2 x - a/2 x + (a/2)^2 = x^2 - a x + (a/2)^2, what we originally had.

  5. AccessDenied
    • 3 years ago
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    * and similarly for y's perfect square.

  6. artofspeed
    • 3 years ago
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    isn't x^2 - 2a x + a^2 = (x - a)^2 ?

  7. AccessDenied
    • 3 years ago
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    Yes, that is correct. However, we had 'x^2 - ax' in our original equation, and added on a (a/2)^2 at the end. :)

  8. AccessDenied
    • 3 years ago
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    Like, we were aiming for a form of x^2 + 2w x + w^2 = (x + w)^2 when we added that (a/2)^2. Let w = -a/2, and we have: x^2 + 2(-a/2) x + (-a/2)^2 = (x - a/2)^2 x^2 - ax + (a/2)^2 = (x - a/2)^2 If that makes more sense.

  9. artofspeed
    • 3 years ago
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    o-o so how can i improve my answers

  10. AccessDenied
    • 3 years ago
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    Instead of (x - a)^2 + (y + b)^2 = ... (the rest I believe is correct so i won't copy it), we'd have (x - a/2)^2 + (y + b/2)^2 = ... (same thing)

  11. artofspeed
    • 3 years ago
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    so instead of writing \[(x-a)^2 + (y+b)^2 = (a/2)^2 + (b/2)^2\] what do you suggest o-o

  12. AccessDenied
    • 3 years ago
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    \[ (x - \frac{a}{2})^2 + (y + \frac{b}{2})^2 = (a/2)^2 + (b/2)^2 \] If it doesn't make sense where those come from, you can expand it out to prove that this is our original problem and that yours doesn't quite come out to be the same. I think its just an error with how you originally completed the square; you added the right term, but you collected it into (x - a)^2 and (y + b)^2, which wouldn't be the same thing..

  13. artofspeed
    • 3 years ago
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    ohhhh ok yea i got it.. sorry for being dumb lol

  14. AccessDenied
    • 3 years ago
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    It's fine. We learn best from mistakes. :) So, from there, you can pull the information pretty easily. I may recommend a small detail on the radius: when you take the square root, you could factor out that 1/2^2 from (a/2)^2 + (b/2)^2 and pull it out as a 1/2. It'd look nicer, I think. :P 1/2 sqrt(a^2 + b^2)

  15. artofspeed
    • 3 years ago
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    oh yea true

  16. artofspeed
    • 3 years ago
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    thx man

  17. AccessDenied
    • 3 years ago
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    You're welcome! :)

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