# the equation of circle in complex plane passing through three points

I need hints on this question

Q.1 Show that the equation of circle passing through three points $z_1, z_2$ and $z_3$ is given by $$\displaystyle \frac{(z-z_1)/(z-z_2)}{(z_3-z_1)/(z_3-z_2)} = \frac{(\bar z-\bar z_1)/(\bar z-\bar z_2)}{(\bar z_3-\bar z_1)/(\bar z_3-\bar z_2)}$$

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What have you tried so far? Where are you stuck? Show some your work so far...I think you also must add the condition that the points $\,z_i\,$ aren't collinear, though the very equation makes this clear as otherwise...? – DonAntonio Feb 23 '13 at 12:03
Do you know something about the functions of the form $z\mapsto \frac{az+b}{cz+d}$? – Hagen von Eitzen Feb 23 '13 at 12:40
@HagenvonEitzen No, could you give me some information on that? – hasExams Feb 23 '13 at 15:01
@DonAntonio So far I know the equation of circle in complex plane is $|z-z_1| = r$ and $az\bar z + b \bar z + \bar b z + c = 0$ and I know how to find the circle if three coordinates are given in xy plane – hasExams Feb 23 '13 at 15:04
About what Hagen wrote: google "fractional transformations" and/or "Moebius transformations" – DonAntonio Feb 23 '13 at 15:55

In $\mathbb{R}^2$, the circle passing through $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ passes satisfies a determinant:

$\left| \begin{array}{cccc} x^2 + y^2 & x & y & 1 \\\\ x^2_1 + y^2_1 & x_1 & y_1 & 1 \\\\ x^2_2 + y^2_2 & x_2 & y_2 & 1 \\\\ x^2_3 + y^2_3 & x_3 & y_3 & 1 \\\\\end{array} \right| = 0$

This is considered a variant of Appolonius' problem of finding a circle tangent to 3 different circles. Other problems are limiting cases, e.g.

We can get a determinant formula for a circle with complex entries by setting $x = \frac{1}{2}(z + \overline{z}),\, y = \frac{1}{2}(z - \overline{z})\, x^2+y^2 = z \overline{z} = |z|^2$:

$\left| \begin{array}{cccc} |z\;|^2 & z & \overline{z} & 1 \\\\ |z_1|^2 & z_1 & \overline{z}_1 & 1 \\\\ |z_2|^2 & z_2 & \overline{z}_2 & 1 \\\\ |z_3|^2 & z_3 & \overline{z}_3 & 1 \\\\ \end{array} \right| = 0$

This analytic formula hides the symmetries of the group of Möbius transformations [video] acting here.

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