Show that the equation delta zbar z+abar z+bar az+γ=0 (with…

Show that the equation $\ \delta z\bar z+a\bar z+\bar az+\gamma=0\ $ (with $\delta,\gamma\in\mathbb R$, $a\in\mathbb C$) represents: $1^\circ$ a circle not through the origin if $\delta\neq0$, $\gamma\neq0$, $\gamma\delta-|a|^2<0$; $2^\circ$ a circle through the origin if $\delta\neq0$, $\gamma=0$; $3^\circ$ a line through the origin if $\delta=\gamma=0$. Conversely, show every circle or line can be written in this form.

Step 1 · Reduce to Cartesian form

With $z=x+iy$, $a=\alpha+i\beta$: $\ z\bar z=x^2+y^2$ and $a\bar z+\bar az=2\text{Re}(a\bar z)=2(\alpha x+\beta y)$. The equation becomes:

$$ \delta(x^2+y^2)+2(\alpha x+\beta y)+\gamma=0 $$

Part $1^\circ$ · Circle not through the origin ($\delta\neq0,\ \gamma\neq0,\ \gamma\delta<|a|^2$)

Dividing by $\delta$ and completing the square:

$$ \left(x+\frac\alpha\delta\right)^2+\left(y+\frac\beta\delta\right)^2=\frac{\alpha^2+\beta^2-\gamma\delta}{\delta^2}=\frac{|a|^2-\gamma\delta}{\delta^2} $$

Since $\gamma\delta-|a|^2<0$, the right side is a positive quantity — a genuine circle, with center $-\dfrac a\delta$ and radius $\dfrac{\sqrt{|a|^2-\gamma\delta}}{|\delta|}$. Setting $z=0$ in the original equation gives $\gamma\neq0$, so the origin is not on the circle.

Part $2^\circ$ · Circle through the origin ($\delta\neq0,\ \gamma=0$)

The same completion of the square gives $\left(x+\frac\alpha\delta\right)^2+\left(y+\frac\beta\delta\right)^2=\dfrac{|a|^2}{\delta^2}$ — a circle of center $-\frac a\delta$, radius $\frac{|a|}{|\delta|}$. Setting $z=0$ now gives $0$, so the origin is on the circle.

Part $3^\circ$ · Line through the origin ($\delta=\gamma=0$)

The equation collapses to $\ a\bar z+\bar az=0\ \Longleftrightarrow\ \alpha x+\beta y=0$, a line through the origin (with direction perpendicular to $(\alpha,\beta)$).

Converse · Every circle is of this form

A circle $(x-h)^2+(y-k)^2=r^2$ expands to $x^2+y^2-2hx-2ky+(h^2+k^2-r^2)=0$. Taking $\delta=1$, $a=-h-ik$, $\gamma=h^2+k^2-r^2$ recovers exactly this equation, since $a\bar z+\bar az=2(-hx-ky)$.

Converse · Every line is of this form

A line $Ax+By+C=0$ ($A,B,C\in\mathbb R$) is recovered with $\delta=0$, $a=\frac{A-iB}2$, $\gamma=C$, since then $a\bar z+\bar az=2\left(\frac A2x+\frac B2y\right)=Ax+By$.

Verification

Example: circle $x^2+y^2-2x=0$ (through the origin, center $(1,0)$, radius $1$). Here $\delta=1$, comparing $-2hx-2ky=-2x\Rightarrow h=1,k=0$, so $a=-1$, $\gamma=1+0-1=0$ ✓ matches Part $2^\circ$. And indeed $\gamma=0$ as required.