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

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AccessDeniedBest ResponseYou've already chosen the best response.1
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?
 one year ago

artofspeedBest ResponseYou've already chosen the best response.0
i got (xa)^2 + (y+b)^2 = (a/2)^2 + (b/2)^2
 one year ago

artofspeedBest ResponseYou've already chosen the best response.0
so the radius must be \[\sqrt{(a/2)^2 + (b/2)^2}\] and center is (a, b) right?
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
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.
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
* and similarly for y's perfect square.
 one year ago

artofspeedBest ResponseYou've already chosen the best response.0
isn't x^2  2a x + a^2 = (x  a)^2 ?
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
Yes, that is correct. However, we had 'x^2  ax' in our original equation, and added on a (a/2)^2 at the end. :)
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
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.
 one year ago

artofspeedBest ResponseYou've already chosen the best response.0
oo so how can i improve my answers
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
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)
 one year ago

artofspeedBest ResponseYou've already chosen the best response.0
so instead of writing \[(xa)^2 + (y+b)^2 = (a/2)^2 + (b/2)^2\] what do you suggest oo
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
\[ (x  \frac{a}{2})^2 + (y + \frac{b}{2})^2 = (a/2)^2 + (b/2)^2 \] 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..
 one year ago

artofspeedBest ResponseYou've already chosen the best response.0
ohhhh ok yea i got it.. sorry for being dumb lol
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
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)
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
You're welcome! :)
 one year ago
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