Answer
1. $(x+\frac{a}{2})^2+(y+\frac{b}{2})^2=\frac{a^2+b^2-4c}{4}$
2. when $a^2+b^2-4c\gt0$, the equation represents a circle with a center at $(-\frac{a}{2},-\frac{b}{2})$
and a radius of $r=\sqrt {\frac{a^2+b^2-4c}{4}}$
3. when $a^2+b^2-4c=0$, the equation represents a point $(-\frac{a}{2},-\frac{b}{2})$;
4. when $a^2+b^2-4c\lt0$, the equation represents an empty set.
Work Step by Step
1. Given $x^2+ax+y^2+by+c=0$, we have
$x^2+ax+(\frac{a}{2})^2+y^2+by+(\frac{b}{2})^2+c=0++(\frac{a}{2})^2+(\frac{b}{2})^2$,
$(x+\frac{a}{2})^2+(y+\frac{b}{2})^2=\frac{a^2}{4}+\frac{b^2}{4}-c$ hence we have a simplified form
$(x+\frac{a}{2})^2+(y+\frac{b}{2})^2=\frac{a^2+b^2-4c}{4}$
2. when $a^2+b^2-4c\gt0$, the equation represents a circle with a center at $(-\frac{a}{2},-\frac{b}{2})$
and a radius of $r=\sqrt {\frac{a^2+b^2-4c}{4}}$
3. when $a^2+b^2-4c=0$, the equation represents a point $(-\frac{a}{2},-\frac{b}{2})$;
4. when $a^2+b^2-4c\lt0$, the equation represents an empty set.
