Circle equation coefficient conditions

[RChristenk]

Homework Statement

Under what conditions on the coefficients $a,b,c$ does the equation $x^2+y^2+ax+by+c=0$ represent a circle? When that condition is satisfied, find the center and radius of the circle.

Relevant Equations

Basic form of circle equation

$x^2+ax+y^2+by=-c$

$\Leftrightarrow (x+\dfrac{a}{2})^2+(y+\dfrac{b}{2})^2=-c+\dfrac{a^2}{4}+\dfrac{b^2}{4}$

The conditions under which the coefficients of this equation makes a circle:

$-c+\dfrac{a^2+b^2}{4}>0$

$\Leftrightarrow 4c < a^2+b^2$

Center of circle: $(-\dfrac{a}{2}, -\dfrac{b}{2})$

Radius of circle: $\sqrt{\dfrac{a^2+b^2-4c}{4}} \Rightarrow \dfrac{\sqrt{a^2+b^2-4c}}{2}$

The question I have is the solution gives the radius of the circle as: $\dfrac{\sqrt{a^2+b^2-4ac}}{2}$. Where did that extra $a$ in $-4ac$ come from?

 

[fresh_42]

I'd say from a typo or a reflex writing the discriminant of $ax^2+bx+c=0$ as $b^2-4ac.$

Since the problem statement is symmetric in $(x,a),(y,b)$ there is no reason for a symmetry break in the root.

 

[RChristenk]

I'd say from a typo or a reflex writing the discriminant of $ax^2+bx+c=0$ as $b^2-4ac.$

Since the problem statement is symmetric in $(x,a),(y,b)$ there is no reason for a symmetry break in the root.

Yes I think the solution is a typo too. I can see the equation is symmetric about the origin, but what do you mean the term $-4ac$ breaks the symmetry in the root? Thanks.

 

[fresh_42]

Yes I think the solution is a typo too. I can see the equation is symmetric about the origin, but what do you mean the term $-4ac$ breaks the symmetry in the root? Thanks.

If the term under the root were $a^2+b^2-4ac$ then $a$ would be distinguishable from $b$. But the original expression $x^2+ax+y^2+by+c=0$ makes no difference between $(x,a)$ and $(y,b)$. We could exchange the two pairs without changing anything. But if we exchange them in $a^2+b^2-4ac$ we would get $a^2+b^2-4bc$ which is not the same.