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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?

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  • $\begingroup$ 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? $\endgroup$ Feb 13, 2019 at 17:35

2 Answers 2

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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 $z_0$

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  • $\begingroup$ But is $|z-z_0|^2=r^2$ still an equation of a circle? $\endgroup$
    – mavavilj
    Jan 19, 2016 at 6:10
  • $\begingroup$ Yes it is a circle $\endgroup$ Jan 19, 2016 at 6:46
  • $\begingroup$ @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 $z_j = x_j+\sqrt{-1}y_j$, $|z_2 - z_1| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = d((x_1,y_1),(x_2,y_2))$. $\endgroup$ Feb 13, 2019 at 17:34
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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

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

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