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artofspeed
 3 years ago
Find the center and radius of this equation of circle x^2+y^2ax+by=0
artofspeed
 3 years ago
Find the center and radius of this equation of circle x^2+y^2ax+by=0

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AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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?

artofspeed
 3 years ago
Best ResponseYou've already chosen the best response.0i got (xa)^2 + (y+b)^2 = (a/2)^2 + (b/2)^2

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

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1I 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.

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1* and similarly for y's perfect square.

artofspeed
 3 years ago
Best ResponseYou've already chosen the best response.0isn't x^2  2a x + a^2 = (x  a)^2 ?

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

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Like, 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.

artofspeed
 3 years ago
Best ResponseYou've already chosen the best response.0oo so how can i improve my answers

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Instead 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)

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

AccessDenied
 3 years ago
Best 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..

artofspeed
 3 years ago
Best ResponseYou've already chosen the best response.0ohhhh ok yea i got it.. sorry for being dumb lol

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1It'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)

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1You're welcome! :)
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