# Showing that $\lvert z-w_0 \rvert = r$ is equivalent to $\alpha {\lvert z \rvert}^2 + \beta Re(z) + \gamma Im(z) - \delta = 0$

I'm having difficulties trying to solve this problem :

The equation $\lvert z-w_0 \rvert = r$ represent the circle in $\Bbb C$ with center $w_0$ and radius $r$.

Show that the above equation is equivalent to

$$\alpha {\lvert z \rvert}^2 + \beta Re(z) + \gamma Im(z) - \delta = 0, \quad \alpha, \beta, \gamma,\delta \in \mathbb{R}, \alpha \neq 0$$

Let's assume $w_0 = (x_0, y_0)$

If $\lvert z-w_0 \rvert = r$ then $(x - x_0)^2 + (y-y_0)^2 = r^2$

$\Rightarrow x^2 - 2{x_0}x + {x_0}^2 + y^2 - 2{y_0}y + {y_0}^2 = r^2$

$\Rightarrow {\lvert z \rvert}^2 + (-2x_0)Re(z) + (-2y_0)Im(z) + ({x_0}^2 + {y_0}^2 + r^2) = 0$

How do I continue from here? Is the demonstration complete?

• Should be. Set $\alpha=1$, $\beta=-2x_0$, $\gamma=-2y_0$, and $\delta=-({x_0}^2+{y_0}^2+r^2)$. – Blake Oct 19 '15 at 21:24

${\lvert z \rvert}^2 + (-2x_0)Re(z) + (-2y_0)Im(z) + ({x_0}^2 + {y_0}^2 - r^2) = 0$ but you have done the work correctly. Simply you have found that: $$\alpha=1 \qquad \beta=-2x_0 \qquad \gamma= -2 y_0 \qquad \delta=x_0^2+y_0^2-r^2$$
(and note that, using $z=(x,y)$, is a well known result in elemental analytic geometry).