Show that the equation represents a circle by rewriting it in standard form, and find the center and the radius of the circle.

Hi sally1,

$$\boxed{x^2+y^2-6x+6y+9=0}$$

$$compare: ax^2+2bxy+cy^2+2dx+2ey+f=0$$

$$b=0 \quad and \quad a=c \quad \Rightarrow \textcolor[rgb]{1,0,0}{\mbox{this is a circle}}$$

$$\mbox{compare for a circle}: x^2+y^2+2mx+2ny+q=0$$

$$2m=-6 \quad and \quad 2n=6 \\ \Rightarrow \\ m=-3 \\ n=3\\ q=9$$

$$\\radius: R = \sqrt{m^2+n^2-q}=\sqrt{(-3)^2+3^2-9}=\sqrt{9}=3\\ Center: x_0=-m=3 \quad and \quad y_0=-n=-3$$

radius = 3

center (3, -3)

$$\boxed{x^2+y^2+2mx+2ny+q=0}$$

Find the standard form:

$$\\x^2+2mx+y^2+2ny=-q\\ x^2+\textcolor[rgb]{1,0,0}{2m}x+y^2+\textcolor[rgb]{0,0,1}{2n}y=-q\\ \underbrace{x^2+\textcolor[rgb]{1,0,0}{2m}x+ (\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2 }_{ (x+m)^2} + \underbrace{y^2+\textcolor[rgb]{0,0,1}{2n}y+( \frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2}_{ (y+n)^2} =-q +(\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2 +(\frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2\\ (x+m)^2+(y+n)^2=m^2+n^2-q\\ \boxed{ \Rightarrow x_0 = -m \qquad y_0=-n \qquad R = \sqrt{m^2+n^2-q}}$$

Hey, sally1, look at this one I just did for you.....

http://web2.0calc.com/questions/show-that-the-equation-represents-a-circle-by-rewriting-it-in-standard-form-and-find-the-center-and-the-radius-of-the-circle

Are you having trouble with "completing the square??"   That's usually the difficulty with these......but they're really pretty similar problems...let me know if you want to review completing the square and I'll cover that......it's useless to try to write equations in some form if we don't know how to get from "A" to "B"

I haven't seen that one before, heureka....but I like it !!!

I'll have to remember that technique !!

Thumbs Up and Points from me!!

$$\boxed{x^2+y^2+2mx+2ny+q=0}$$

Find the standard form:

$$\\x^2+2mx+y^2+2ny=-q\\ x^2+\textcolor[rgb]{1,0,0}{2m}x+y^2+\textcolor[rgb]{0,0,1}{2n}y=-q\\ \underbrace{x^2+\textcolor[rgb]{1,0,0}{2m}x+ (\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2 }_{ (x+m)^2} + \underbrace{y^2+\textcolor[rgb]{0,0,1}{2n}y+( \frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2}_{ (y+n)^2} =-q +(\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2 +(\frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2\\ (x+m)^2+(y+n)^2=m^2+n^2-q\\ \boxed{ \Rightarrow x_0 = -m \qquad y_0=-n \qquad R = \sqrt{m^2+n^2-q}}$$

Thanks for the further explanation....I was trying to see how you determined the radius and that made it clear !!!

I'm going to look at his one a little more......

I want to see what all the fuss is about

$$x^2+y^2-6x+6y+9=0\\\\ x^2-6x+y^2+6y=-9\\\\ x^2-6x+9+y^2+6y+9=-9+9+9\\\\ (x-3)^2+(y+3)^2=9\\\\$$

Circle centre (3,-3)  radius=3

What was all the big fuss about - did I do something wrong?  I don't think so.

I think you guys are making a mountain out of a mole hill!

I love your new avatar reinout-g!

Thank you