# General Form of equation for a Circle

1. Jul 24, 2007

I was helping my girlfriend earlier with some HW and now I don't have her text (since she is not here) but I have a question regarding circles.
I did not cover a lot of this material in precalc. and only recall the Standard Form of an equation: x^2+y^2=r^2.

So my question is: if given the equation in General Form, and I wish to know the radius, is the following procedure (completing the square) correct?

Given x^2+y^2+ax+by+c=0

x^2+ax+y^2+by+c=o
x^2+ax+(a/2)^2+y^2+by+(b/2)^2+c=(a/2)^2+(b/2)^2
(x+a/2)^2+(y+b/2)^2=(a/2)^2+(b/2)^2-c

so r=sqrt((a/2)^2+(b/2)^2-c)

is this correct?
thanks,
Casey

2. Jul 24, 2007

### CompuChip

Yep, it is correct - and a nice way to do it too.

Another way to derive this is as follows: the equation of a circle with radius r and center point ($x_0$, $y_0$) is $$(x - x_0)^2 + (y - y_0)^2 = r^2$$. Working out the brackets gives $$x^2 - 2 x_0 x + x_0^2 + y^2 - 2 y_0 y + y_0^2 - r^2 = 0$$. Now if we compare to your formula $$x^2 + y^2 + a x + b y + c = 0$$ we read off that
$$a = -2 x_0, b = 2 y_0, c = x_0^2 + y_0^2 - r^2$$
whence solving for r gives
$$r^2 = x_0^2 + y_0^2 - c$$ and you get the same formula (you can substitute -a/2 for $x_0$ and drop the minus since it's squared anyway, and similarly for -b/2).

3. Jul 24, 2007