# Analytic Geometry I - Circle

1. Nov 20, 2008

### kLownn

1. The problem statement, all variables and given/known data
Find the centre and radius of the circle x²+y²-4x+2y+6=0

I have the solution. The circle is no defined because r² = -1 is impossible.
But... how do I even DO that equation to get the answer -1?!

2. Nov 20, 2008

### Dick

You complete the squares so you can write it in the form (x+a)^2+(y+b)^2=c. Do you know how to do that?

3. Nov 20, 2008

### kLownn

I learned how to do that while I was in school.. I forget how to do it now.. >_<

4. Nov 20, 2008

### Dick

Take the x part. You've got x^2-4x. If I add something to that it will become a perfect square of the form (x-a)^2. What's 'a'? What do you have to add?

5. Nov 20, 2008

### kLownn

Oh! Is "a" 2x?
(x-2x)² ?

6. Nov 20, 2008

### Dick

No, no. (x-a)^2=x^2-2ax+a^2, yes? If you match that up with x^2-4x, the 4 must be the 2a, as I see it. Think back to when you did this before.

7. Nov 20, 2008

### kLownn

Ohh, I see now.. so I just do the same thing for y?

8. Nov 20, 2008

### Dick

Sure.

9. Nov 20, 2008

### kLownn

Thank you so much! :)

10. Nov 21, 2008

### icystrike

it can actually be express as x²+y²+2fx+2gy+c=0

whrby C(-f,-g) and radius is $$\sqrt{g²+f²-c}$$

11. Nov 21, 2008

### HallsofIvy

Staff Emeritus
No, it isn't. It is much better to actually do the "complete the square" rather than memorize formulas: so you don't make silly mistakes like that.

12. Nov 21, 2008

### icystrike

yep (: noted.