# The equation of a circle on a complex plane?

The equation of a circle $|z-z_0|=r$ in a complex plane has (among others) the form:

$$z\overline{z}+\overline{b}z+b\overline{z}+c=0$$ where $b=-z_0 \in \mathbb{C}$.

What I'd like to understand is, why is it so?

• It occurs to me that your question is possibly unclear. Do you want to know why $|z-z_0| - r = 0$ can expand into this form, or why it defines a circle? Feb 13, 2019 at 17:35

It's because you can write the original form as $$|z-z_0|^2=r^2$$ and $$|z-z_0|^2=(z-z_0)(\overline{z-z_0})=(z-z_0)(\overline{z}-\overline{z_0})$$ Now substitute for $z_0$

• But is $|z-z_0|^2=r^2$ still an equation of a circle? Jan 19, 2016 at 6:10
• Yes it is a circle Jan 19, 2016 at 6:46
• @mavavilj A circle with center $z_0$ and radius $r$ is the set of all points $z$ with distance $r = d(z,z_0)$. Now recall that for $z_j = x_j+\sqrt{-1}y_j$, $|z_2 - z_1| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = d((x_1,y_1),(x_2,y_2))$. Feb 13, 2019 at 17:34

A particularly simple equation is that of a circle:

$$\{z: |z - a| = r\}$$

is the circle with radius $$r$$ and center $$a$$. By squaring that equation we obtain

$$(z - a)(z' - a') = r²$$

or

$$zz' - (za' + z'a) + (aa' - r²) = 0$$

and finally

$$zz' - (za' + z'a) + s = 0$$

where $$s$$ is a real number. The circle is centered at a and has the radius $$r = \sqrt{a}a' - s$$, provided the root is real.

This representation of the circle is more convenient in some respects. For example, we may immediately check that the transformation $$w = f(z) = \frac{1}{z}$$ maps circles onto circles. Indeed, substituting $$z = \frac{1}{w}$$ we get

$$\frac{1}{w} \times \frac{1}{w'} - (\frac{a'}{w} + \frac{a}{w'}) + s = 0$$

which, if multiplied by $$ww'$$, leads to

$$ww' - (wb' + w'b) + t = 0$$

where $$b = \frac{a'}{s}$$ and $$t = \frac{1}{s}$$, an equation in the same form.

Letting $$a = \alpha + i\beta$$ yields yet another form of essentially same equation:

$$zz' - \alpha(z + z') - i\beta(z - z') + s = 0$$

where $$\alpha$$ and $$\beta$$ are both real. Yet the most general form of the equation is this

$$Azz' + Bz + Cz' + D = 0$$