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My book gives the equation $z\bar z + a \bar z + \bar a z+ b = 0$ for general equation of circle with centre -a and radius $\sqrt {a \bar a - b}$. I can understand that general equation of a circle can be $|z-z_0|=r$ or $z \bar z - z_0 \bar z -\bar z_0 z + z_0 \bar z_0 - r^2 = 0$, but how does that give rise to this equation? I am referring to the formula list in the Arihant 39 years JEE solved papers.

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    $\begingroup$ Compare coefficients of $z$ and $\bar z$ $\endgroup$ Commented Jun 2, 2018 at 12:44

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Note that\begin{align}z\overline z+a\overline z+\overline az+b=0&\iff(z+a)\overline{(z+a)}=a\overline a-b\\&\iff|z+a|^2=a\overline a-b.\\&\iff\bigl|z-(-a)\bigr|^2=a\overline a-b.\end{align}

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In

$$ Z \bar Z = r^2 $$

making $Z = z + a$

$$ (z+a)(\bar z + \bar a) = r^2 $$

or

$$ z\bar z+z \bar a+ \bar z a + a\bar a -r^2 = 0 $$

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