Equation of a Circle in Standard Form

(i) Centre at the origin. Let the centre be $C(0,0)$, the radius $r$, and $P(x,y)$ any point on the circle. Since $CP=r$ for every position of $P$,

This is the equation of a circle with centre the origin and radius $r$.

(ii) Centre at $(h,k)$. With centre $C(h,k)$ and radius $r$, the same distance condition $CP=r$ gives the standard (centre–radius) form

Expanding, $x^2+y^2-2hx-2ky+(h^2+k^2-r^2)=0$. Writing $g=-h$, $f=-k$, $c=h^2+k^2-r^2$ turns this into the general form

whose centre is $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$ (found by completing the square back to the standard form). This equation has three tell-tale features: it is second degree in $x,y$; the coefficients of $x^2$ and $y^2$ are equal and nonzero; and there is no $xy$ term.

Converse (worth proving, since it is used constantly to recognise a circle at a glance). Any equation $ax^2+ay^2+2g'x+2f'y+c'=0$ ($a\ne0$) with those three features can be divided by $a$ and completed to the square exactly as above, landing back in the standard form — so it always represents a circle, with centre $\left(-\frac{g'}a,-\frac{f'}a\right)$ and radius $\sqrt{\left(\frac{g'}a\right)^2+\left(\frac{f'}a\right)^2-\frac{c'}a}$.

Theorem 5.1 (family of circles through a line–circle intersection). For the circle $S:x^2+y^2+2gx+2fy+c=0$ and the line $L:lx+my+n=0$, the equation

$S+\lambda L=0,\qquad \lambda\in\mathbb R,$ …