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Find the center and radius of the circle which circumscribes the triangle with (complex) vertices $a_1,a_2,a_3$. Express the result in symmetric form.

I'm not sure where to start in this question. The circumcenter is the intersection of the perpendicular bisectors, but I don't know how to calculate the line equation of the perpendicular bisectors yet. As for the radius, after we find the circumcenter $c$ we can calculate it using $|c-a_1|$ (or $|c-a_2|$ or $|c-a_3|$; all three should be equal.) So how can I calculate the circumcenter?

Paul S.
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  • This is a standard result of complex numbers in geometry. You should learn the approaches and derive the answer. – Calvin Lin Sep 01 '13 at 15:01
  • @CalvinLin How hard are the "approaches"? My book just introduces complex numbers and talks about radius and arguments, but nothing further beyond that. Then it has this as an exercise. – Paul S. Sep 01 '13 at 15:05
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    The perpendicular bisector of the side with the vertices $a_1$ and $a_2$ is $$\frac{a_1+a_2}{2} + i t (a_2-a_1), \quad t \in \mathbb{R}.$$ – njguliyev Sep 01 '13 at 15:13
  • @njguliyev Okay I see where that comes from. So we must solve the equation $$\frac{a_1+a_2}{2}+it_1(a_2-a_1)=\frac{a_1+a_3}{2}+it_2(a_3-a_1)$$, right? What can we do since there are two variables $t_1,t_2$? – Paul S. Sep 01 '13 at 15:29
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    You can find $t$ from $$\left(\frac{a_1+a_2}{2} + i t (a_2-a_1)-a_3\right)\overline{\left(\frac{a_1+a_2}{2} + i t (a_2-a_1)-a_3\right)} = |a_2-a_1|^2 (1/4 + t^2).$$ – njguliyev Sep 01 '13 at 15:42
  • @njguliyev This time I don't see where that comes from... – Paul S. Sep 01 '13 at 17:25
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    The left-hand side is the square of the distance between the point on the perpendicular bisector and $a_3$, the right-hand side is the square of the distance between the point on the perpendicular bisector and $a_1$. – njguliyev Sep 01 '13 at 18:27

2 Answers2

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Let $R$ be the radius of the circumcircle of $a_1,a_2,a_3$ then the circumcenter $z$ is defined by the condition $|z-a_1|=|z-a_2|=|z-a_3|=R\,$. Expanding $R^2=|z-a_1|^2=(z-a_1)(\bar z - \overline{a_1})$ and similar for $a_2,a_3$ gives the system:

$$ \begin{cases} \begin{align} \overline{a_1}\,z + a_1\,\bar z + (R^2 - |z|^2) & = |a_1|^2 \\ \overline{a_2}\,z + a_2\,\bar z + (R^2 - |z|^2) & = |a_2|^2 \\ \overline{a_3}\,z + a_3\,\bar z + (R^2 - |z|^2) & = |a_3|^2 \end{align} \end{cases} $$

Eliminating $\bar z$ and $(R^2 - |z|^2)$ between the equations amounts to solving it as a linear system in $z$,$\bar z$ and $(R^2-|z|^2)$ which, by Cramer's rule, gives:

$$ z\;=\;\frac{\left| \begin{array}{ccc} |a_1|^2 & a_1 & -1 \\ |a_2|^2 & a_2 & -1 \\ |a_3|^2 & a_3 & -1 \end{array} \right|} {\left| \begin{array}{ccc} \overline{a_1} & a_1 & -1 \\ \overline{a_2} & a_2 & -1 \\ \overline{a_3} & a_3 & -1 \end{array} \right|} \;=\;\frac{\left| \begin{array}{ccc} a_1 & |a_1|^2 & 1 \\ a_2 & |a_2|^2 & 1 \\ a_3 & |a_3|^2 & 1 \end{array} \right|} {\left| \begin{array}{ccc} a_1 & \overline{a_1} & 1 \\ a_2 & \overline{a_2} & 1 \\ a_3 & \overline{a_3} & 1 \end{array} \right|} $$

(The formula itself has been posted in another answer here. The above is a derivation thereof.)

Community
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dxiv
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Here is a way of getting started. It might not be the best way, but it makes significant progress. Let $c$ be the centre of the circle and $r$ (a real number) the radius. You can write an equation for the square of the radius as $$(a_1-c)(\bar{a}_1-\bar c)=r^2=|a_1|^2+|c|^2-c\bar a_1-a_1\bar c$$

[note this is just the cosine rule for triangles in different notation]

You can write two other equations like this, and use them to eliminate $r^2$, $|c|^2$ (which appear together) and $\bar c$ so that you have a linear equation for $c$ in terms of things you know.

Mark Bennet
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    Thanks, Mark. I calculated using your method and found that $$c=\frac{a_2(|a_1|^2-|a_3|^2)+a_3(|a_2|^2-|a_1|^2)+a_1(|a_3|^2-|a_2|^2)}{a_1\overline{a_3}+a_3\overline{a_2}+a_2\overline{a_1}-a_1\overline{a_2}-a_2\overline{a_3}-a_3\overline{a_1}}$$ This is quite a big mess! What would be the best way to calculate $r$ from here? – Paul S. Sep 01 '13 at 17:38
  • We have $$a_1-c=\dfrac{a_1^2\overline{a_3}+a_1a_3\overline{a_2}+a_2|a_3|^2+|a_2|^2a_1-a_1^2\overline{a_2}-a_1a_2\overline{a_3}-|a_2|^2a_3-|a_3|^2a_1}{a_1\overline{a_3}+a_3\overline{a_2}+a_2\overline{a_1}-a_1\overline{a_2}-a_2\overline{a_3}-a_3\overline{a_1}}$$ That is huge! – Paul S. Sep 01 '13 at 17:43
  • I certainly don't want to multiply out $(a_1-c)\overline{(a_1-c)}$! – Paul S. Sep 01 '13 at 17:45
  • @PaulS. I got $(a_1-a_2)(a_2-a_3)(a_3-a_1)=((a_1\bar a_2- \bar a_1 a_2)+(a_2\bar a_3-\bar a_2 a_3)+(a_3\bar a_1-\bar a_3a_1))c$ - which at least shows some sensible symmetry. – Mark Bennet Sep 01 '13 at 18:47
  • @PaulS. Given this uses the cosine rule, and the long form of the sine rule is $\cfrac a{\sin A}=2r$ it would be possible to compute from that - $A$ comes from the arguments, and the length of the side can be computed. I am not saying these are the best ways to progress, just that they are ways of making progress which ultimately work. – Mark Bennet Sep 01 '13 at 18:50