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

1. AccessDenied

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

i got (x-a)^2 + (y+b)^2 = (a/2)^2 + (b/2)^2

3. artofspeed

so the radius must be $\sqrt{(a/2)^2 + (b/2)^2}$ and center is (a, -b) right?

4. AccessDenied

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

* and similarly for y's perfect square.

6. artofspeed

isn't x^2 - 2a x + a^2 = (x - a)^2 ?

7. AccessDenied

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

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

o-o so how can i improve my answers

10. AccessDenied

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

so instead of writing $(x-a)^2 + (y+b)^2 = (a/2)^2 + (b/2)^2$ what do you suggest o-o

12. AccessDenied

$(x - \frac{a}{2})^2 + (y + \frac{b}{2})^2 = (a/2)^2 + (b/2)^2$ 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

ohhhh ok yea i got it.. sorry for being dumb lol

14. AccessDenied

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

oh yea true

16. artofspeed

thx man

17. AccessDenied

You're welcome! :)