Question

Show that X^2 + Y^2 + Dx + Ey + F=0 represents a circle of positive radius if and only if D^2 + E^2 - 4F > 0.

Answer 1

Show that $x^2+y^2+Dx+Ey+F=0$ represents a circle of positive radius if and only if $D^2 + E^2 - 4F > 0$ .

Solution:

The equation of the circle can be written as

$(x-a)^2+(y-b)^2=R^2$ , where $(a, b)$ is a centre of the circle, R is a radius of the circle.

Let's transform our equation to the form of the equation of the circle:

$x^2+y^2+Dx+Ey+F=x^2+2\frac D2x+\frac{D^2}{4}+$$y^2+2\frac E2y+\frac{E^2}{4}-\frac{D^2}{4}-\frac{E^2}{4}+F=0$

$(x+\frac D2)^2+(y+\frac E2)^2-\frac{D^2}{4}-\frac{E^2}{4}+F=0$

$(x+\frac D2)^2+(y+\frac E2)^2=\frac{D^2}{4}+\frac{E^2}{4}-F$ (*)

Compare (*) and the equation of the circle. Equation (*) represents a circle if and only if $\frac{D^2}{4}+\frac{E^2}{4}-F=R^2$ . $R^2$ is always a positive value so we can conclude

$\frac{D^2}{4}+\frac{E^2}{4}-F>0$

or

$D^2+E^2-4F>0$