What is the standard equation of a circle?

Question

michele_laino · Answer

here is the standard equation of a circle:
$$\Large  {x^2} + {y^2} + ax + by + c = 0$$
where, a, b, and c are real coefficient, which satisfy that condition:
 $$\Large  \frac{{{a^2}}}{4} + \frac{{{b^2}}}{4} - c > 0$$

michele_laino · Answer

coefficients*

anonymous · Answer

I always thought the "standard form" of the equation of a circle was$$\left( x-a \right)^{2} + \left( y-b \right)^{2} = r ^{2}$$where the center of the circle is at (a, b) and the radius is r.

anonymous · Answer

Haha, me too...

freckles · Answer

I think the first way mentioned by Michele is the general form and the way mentioned by osprey is the standard form. 
Though I'm not sure if these terms hold universal.

anonymous · Answer

@Michele_Laino 's equation results in an ellipse.

anonymous · Answer

Ok, think I got the answer, thanks!

michele_laino · Answer

please, note that, my equation is not the equation of an ellipse @ospreytriple

anonymous · Answer

Let a=4, b=3, and c=1, for example. Graph it and see what you get, @Michele_Laino

michele_laino · Answer

that is a condition, on parameters a, b, and c

anonymous · Answer

???? If the equation truly represents a circle, then the values of a, b, and c shouldn't matter so long as they satisfy the condition you mentioned. It should always produce a circle. And it doesn't.

freckles · Answer

Michele_Laino has written general form for equation @ospreytriple $$x^2+y^2+ax+by+c=0 \ x^2+ax+y^2+by=-c \ (x^2+ax+(\frac{a}{2})^2)+(y^2+by+(\frac{b}{2})^2)=-c +(\frac{a}{2})^2+(\frac{b}{2})^2 \  (x+\frac{a}{2})^2+(y+\frac{b}{2})^2 =\frac{a^2}{4}+\frac{b^2}{4}-c \ \text{ center is } (\frac{-a}{2},\frac{-b}{2}) \ \text{ and radius is } \sqrt{\frac{a^2}{4}+\frac{b^2}{4}-c} \ \text{ this is a circle if } \frac{a^2}{4}+\frac{b^2}{4}-c>0 $$

freckles · Answer

general form for circle *

freckles · Answer

http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%2B4x%2B3y%2B1%3D0 the example you picked shows a circle

anonymous · Answer

Must be something wrong with my version of Autograph as it clearly shows an ellipse. If so, please accept my apologies. I do, however, stand by my "standard form" equation.