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I've been led to believe that for the equation of a circle in a complex plane:

$$z\overline{z}+\overline{B}z+B\overline{z}+c=0$$

adding $a \in \mathbb{R}$ s.t.

$$az\overline{z}+\overline{B}z+B\overline{z}+c=0$$

doesn't change that it's an equation of a circle.

But what's $a$ doing?

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  • $\begingroup$ Changes the center of the circle from $-B$ to $-B/a$ and divides the radius by $\sqrt{a}$. (If $a<0$ and the first equation defines a circle, then the new solution set is empty.) $\endgroup$
    – Did
    Jan 19, 2016 at 12:42
  • $\begingroup$ @Did why does it divide the radius by $\sqrt{a}$? $\endgroup$
    – mavavilj
    Jan 19, 2016 at 14:41
  • $\begingroup$ Compare the radiuses of the circles $z\bar z-1=0$ and $4z\bar z-1=0$. $\endgroup$
    – Did
    Jan 19, 2016 at 14:54

2 Answers 2

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Consider a circle in the complex plane with a clearer representation:

$$|z-d| = r$$

i.e., a circle of radius $r$ centered at $d$. This is equivalent to

$$z \bar{z} - \bar{d} z - d \bar{z} + d \bar{d} - r^2 = 0$$

Now multiply through by a constant $a$ and get

$$a z \bar{z} + \bar{B} z + B \bar{z} +c = 0$$

where $B = -a d$ and $c = a (d \bar{d}-r^2)$. The constant $a$ then simply scales the center and radius with respect to $z$.

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  • $\begingroup$ How do you see that: " The constant a then simply scales the center and radius with respect to z ." $\endgroup$
    – mavavilj
    Jan 19, 2016 at 14:16
  • $\begingroup$ Also, in which direction(s) is the center scaled? $\endgroup$
    – mavavilj
    Jan 19, 2016 at 14:24
  • $\begingroup$ @mavavilj: $a$ is real according to the OP. Thus it moves the center along a line through the origin. $\endgroup$
    – Ron Gordon
    Jan 19, 2016 at 14:28
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    $\begingroup$ @mavavilj: the center - the center and the origin. At this point, are you merely curious or are you testing me? $\endgroup$
    – Ron Gordon
    Jan 19, 2016 at 14:31
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    $\begingroup$ @Did's comment is a succinct statement what what $a$ is "doing." His statement is verified by my algebra. $\endgroup$
    – Ron Gordon
    Jan 19, 2016 at 14:35
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It's doing nothing, provided $a\ne0$, it's just a more general form; just divide by $a$ and you get the previous form.

However, the more general form also includes lines, precisely when $a=0$ (and $B\ne0$).

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