# Finding center and radius of circumcircle

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?

• This is a standard result of complex numbers in geometry. You should learn the approaches and derive the answer. Sep 1, 2013 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. Sep 1, 2013 at 15:05
• The perpendicular bisector of the side with the vertices $a_1$ and $a_2$ is $$\frac{a_1+a_2}{2} + i t (a_2-a_1), \quad t \in \mathbb{R}.$$ Sep 1, 2013 at 15:13
• You can find $t$ from $$\left(\frac{a_1+a_2}{2} + i t (a_2-a_1)-a_3\right)\overline{\left(\frac{a_1+a_2}{2} + i t (a_2-a_1)-a_3\right)} = |a_2-a_1|^2 (1/4 + t^2).$$ Sep 1, 2013 at 15:42
• 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 $a_1$. Sep 1, 2013 at 18:27

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.

• 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? Sep 1, 2013 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! Sep 1, 2013 at 17:43
• I certainly don't want to multiply out $(a_1-c)\overline{(a_1-c)}$! Sep 1, 2013 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. Sep 1, 2013 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. Sep 1, 2013 at 18:50

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.)