General Equation of a Circle

General Equation of a Circle

From centre-radius form to a standard second-degree pattern. Start from the centre-radius form with centre $(h,k)$ and radius $r$:

Expanding,

Now compare this with the form $x^2+y^2+2gx+2fy+c=0$. Matching coefficients term by term:

so $(h,k)=(-g,-f)$ and $r^2=h^2+k^2-c=g^2+f^2-c$, i.e. $r=\sqrt{g^2+f^2-c}$ (this needs $g^2+f^2-c\ge0$ for $r$ to be a real, meaningful radius). This proves:

The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$, with centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$ (provided $g^2+f^2-c>0$).

Reading off a circle from its general equation. Given any equation of this shape, you can find $g$ by halving the coefficient of $x$, $f$ by halving the coefficient of $y$, and $c$ as the constant term — then centre and radius follow immediately from the boxed formulas above.

Guided activity — rebuilding the centre-radius form by completing the square. Starting again from $x^2+y^2+2gx+2fy+c=0$, group the $x$-terms and $y$-terms and complete the square on each: $x^2+2gx+g^2$ is $(x+g)^2$, and $y^2+2fy+f^2$ is $(y+f)^2$. Adding and subtracting $g^2+f^2$ to balance the equation:

which is exactly $[x-(-g)]^2+[y-(-f)]^2=\left(\sqrt{g^2+f^2-c}\right)^2$ — the centre-radius form with centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$, confirming the formulas found above by direct algebraic manipulation rather than coefficient matching.

Three cases (Let's Remember). Whether $x^2+y^2+2gx+2fy+c=0$ actually represents a circle in the plane depends entirely on the sign of $g^2+f^2-c$:

1. If $g^2+f^2-c>0$: the equation represents a genuine circle, with a positive real radius.
2. If $g^2+f^2-c=0$: the "radius" is $0$, so the equation represents a single point $(-g,-f)$ — a degenerate circle, the limiting case as the radius shrinks to zero.
3. If $g^2+f^2-c<0$: there is no real point satisfying the equation at all — no radius can be negative, so no circle (real or degenerate) exists in the $xy$-plane.

Recognising the general form. A second-degree equation in $x$ and $y$ represents some circle only if it has no $xy$-term and the coefficients of $x^2$ and $y^2$ are equal (so it can be scaled to have both equal to $1$, matching the pattern above) — and even then, only if $g^2+f^2-c\ge0$.

Solved Example 1 — prove $3x^2+3y^2-6x+4y-1=0$ is a circle; find centre and radius

The leading coefficients are both $3$ (equal, and there is no $xy$ term), so divide the whole equation by $3$ to match the standard pattern:

Comparing with $x^2+y^2+2gx+2fy+c=0$: $2g=-2 \Rightarrow g=-1$; $2f=\frac43 \Rightarrow f=\frac23$; $c=-\frac13$.

Check $g^2+f^2-c$: $(-1)^2+\left(\frac23\right)^2-\left(-\frac13\right)=1+\frac49+\frac13=\frac99+\frac49+\frac39=\frac{16}{9}$, which is positive, so the equation does represent a circle. …