Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.