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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 $$\displaystyle \frac{(z-z_1)/(z-z_2)}{(z_3-z_1)/(z_3-z_2)} = \frac{(\bar z-\bar z_1)/(\bar z-\bar z_2)}{(\bar z_3-\bar z_1)/(\bar z_3-\bar z_2)}$$

hasExams
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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
  • wow!! i didn't know – hasExams Feb 23 '13 at 16:53

2 Answers2

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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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cactus314
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  1. Start with three points a, b and c on the circle.
  2. Define a linear fractional transformation by sending z to the cross ratio (z, a, b, c). Call it f(z).
  3. Note that this is invertible on the riemann sphere.
  4. For any point z, look at the complex conjugate of f(z), and apply the inverse of f to that point. Let the resultant point be w. Then we say that z and w are symmetric about your circle.
  5. Try to prove that f sends your circle onto the real axis along with the point at infinity.
  6. Conclude by noting that z lies on the circle iff f(z) is real or the point at infinity.
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