Two formulas for the circumcenter – High School Geometry

Given a triangle $ABC$ , it is possible to draw a special circle that goes through the three vertices $A,B,C$ . That circle is called the circumscribed circle, and its center is called the circumcenter of the triangle. Knowing the circumcenter is crucial to drawing the circumscribed circle, and the standard way of finding the (coordinates of the) circumcenter consists of two main steps:

find the equations of the right bisectors of any two sides;
solve the equations in step 1 to determine their point of intersection, which will then be the desired circumcenter.

In this post, we present two formulas for finding the coordinates of the circumcenter of a triangle, given the coordinates of the vertices of the triangle. Despite this, the standard way of proceeding outlined above is still the preferred strategy.

Formula 1

Maybe this formula executes as fast as the cars in the popular “Formula 1” sport? You be the judge.

Let $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ be the vertices of a $\triangle ABC$ . Then the coordinates of its circumcenter are given by:

$\begin{equation*} \begin{split} x&=\frac{x_1^2(y_2-y_3)+x_2^2(y_3-y_1)+x_3^2(y_1-y_2)-(y_1-y_2)(y_2-y_3)(y_3-y_1)}{2\Big(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\Big)}\\ y&=\frac{y_1^2(x_3-x_2)+y_2^2(x_1-x_3)+y_3^2(x_2-x_1)-(x_3-x_2)(x_2-x_1)(x_1-x_3)}{2\Big(y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)\Big)} \end{split} \end{equation}$

Beautiful, beautiful thing. Notice that the denominators in both expressions are equal (see the exercises at the end), and so it is enough to calculate just one of them to save time (see Example 2 below).

Example 1

Find the circumcenter of $\triangle ABC$ with vertices at $A(0,0),~B(0,3),~C(4,0)$ .

Put $A(0,0)=(x_1,y_1),~B(0,3)=(x_2,y_2),~C(4,0)=(x_3,y_3)$ . Then, “Formula 1” above gives

$\begin{equation*} \begin{split} x&=\frac{0^2(3-0)+0^2(0-0)+4^2(0-3)-(0-3)(3-0)(0-0)}{2\Big(0(3-0)+0(0-0)+4(0-3)\Big)}\\ &=\frac{-48}{-24}\\ &=2\\ &\vdots\cdots\vdots\\ y&=\frac{0(4)+9(-4)+0-0}{2\Big(3(-4)\Big)}\\ &=\frac{-36}{-24}\\ &=\frac{3}{2} \end{split} \end{equation}$

The circumcenter is at $(2,\frac{3}{2})$ . You can verify this by the standard procedure (or, in this case, by observing that the given triangle is a right triangle, and so the circumcenter is at the midpoint of the hypotenuse).

Example 2

Find the circumcenter of a triangle $ABC$ with vertices at $A(-2,1),~B(3,4),~C(6,-5)$ .

Let $A(-2,1)=(x_1,y_1),~B(3,4)=(x_2,y_2),~C(6,-5)=(x_3,y_3)$ . We first calculate the “common denominator” in the expressions for $x$ and $y$ :

$\begin{equation*} \begin{split} \textrm{\textbf{denominator}}&=2\Big(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\Big)\\ &=2\Big(-2(9)+3(-6)+6(-3)\Big)\\ &=2(-54)\\ &=-108 \end{split} \end{equation}$

This will then appear in the denominators for the $x$ and $y$ coordinates of the circumcenter.

The numerator for the $x$ coordinate is: $x_1^2(y_2-y_3)+x_2^2(y_3-y_1)+x_3^2(y_1-y_2)-(y_1-y_2)(y_2-y_3)(y_3-y_1)$ ,
where, by substitution, we obtain

$\begin{equation*} \begin{split} &=4(9)+9(-6)+36(-3)-(-3)(9)(-6)\\ &=36-54-108-162\\ &=-288 \end{split} \end{equation}$

So the $x$ coordinate of the circumcenter is $\frac{-288}{-108}=\frac{8}{3}$ .

Similarly, the numerator for the $y$ coordinate is:
$y_1^2(x_3-x_2)+y_2^2(x_1-x_3)+y_3^2(x_2-x_1)-(x_3-x_2)(x_2-x_1)(x_1-x_3)$ ,
where, by substitution, we obtain:

$\begin{equation*} \begin{split} &=1(3)+16(-8)+25(5)-(3)(-8)(5)\\ &=3-128+125+120\\ &=120 \end{split} \end{equation}$

So the $y$ coordinate of the circumcenter is $\frac{120}{-108}=-\frac{10}{9}$ . Putting the $x$ and $y$ coordinates together, we obtain $\Big(\frac{8}{3},-\frac{10}{9}\Big)$ as the circumcenter.

We’ll redo this example following the regular procedure. In fact, as we mentioned before, the regular procedure is the best when it comes to numerical problems (as it can be somewhat confusing to keep track of the coordinates in the general formula); the general formula comes in handy when deducing general properties.

Example 3

Find the circumcenter of a triangle $ABC$ with vertices at $A(-2,1),~B(3,4),~C(6,-5)$ .

First, a diagram showing the right bisector of $AB$ (dashed line $RP$ ) and the right bisector of $AC$ (dashed line $NS$ ):

Let’s find the equations of these right bisectors.

For the bisector $RP$ of side $AB$ , note that since $R$ is the midpoint of $AB$ , then $R$ is located at $(\frac{1}{2},\frac{5}{2})$ . The slope of $AB$ is $\frac{3}{5}$ , so the slope of $RP$ is $-\frac{5}{3}$ . The equation of $RP$ is then:

$\begin{equation*} \begin{split} y&=mx+b\\ y&=-\frac{5}{3}x+b\\ &\vdots\cdots\vdots\\ \frac{5}{2}&=-\frac{5}{3}\Big(\frac{1}{2}\Big)+b\quad\textrm{using point} R(1/2,5/2)\\ \frac{5}{2}&=-\frac{5}{6}+b\\ \frac{5}{2}+\frac{5}{6}&=b\\ \frac{5\times 3}{2\times 3}+\frac{5}{6}&=b\quad\textrm{common denominator}\\ \frac{15}{6}+\frac{5}{6}&=b\\ \frac{20}{6}&=b\\ \frac{10}{3}&=b\quad\textrm{reduce to lowest terms}\\ &\vdots\cdots\vdots\\ \textrm{So,}~y&=-\frac{5}{3}x+\frac{10}{3} \end{split} \end{equation}$

Next, we find the equation of the bisector $NS$ of side $AC$ . Note that since $N$ is the midpoint of $AC$ , $N$ is located at $(2,-2)$ . The slope of $AC$ is $-\frac{3}{4}$ , so the slope of $NS$ is $\frac{4}{3}$ . Thus, we can write the equation of $NS$ in the form $y=\frac{4}{3}x+b$ , where $b$ can be determined as shown below:

$\begin{equation*} \begin{split} y&=\frac{4}{3}x+b\\ -2&=\frac{4}{3}(2)+b\quad\textrm{using the point}~N(2,-2)\\ -2&=\frac{8}{3}+b\\ -2-\frac{8}{3}&=b\\ \frac{-2\times 3}{1\times 3}-\frac{8}{3}&=b\quad\textrm{common denominator}\\ -\frac{6}{3}-\frac{8}{3}&=b\\ -\frac{14}{3}&=b\\ \vdots\cdots\vdots\\ \textrm{So,}~y&=\frac{4}{3}x-\frac{14}{3} \end{split} \end{equation}$

We now solve the two equations simultaneously using substitution.

$\begin{equation*} \begin{split} y&=-\frac{5}{3}x+\frac{10}{3}\\ y&=\frac{4}{3}x-\frac{14}{3}\\ &\vdots\cdots\vdots\\ -\frac{5}{3}x+\frac{10}{3}&=\frac{4}{3}x-\frac{14}{3}\\ -\frac{5}{3}x-\frac{4}{3}x&=-\frac{14}{3}-\frac{10}{3}\\ -\frac{9}{3}x&=-\frac{24}{3}\\ -3x&=-8\\ x&=\frac{8}{3}\\ &\vdots\cdots\vdots\\ y&=\frac{4}{3}x-\frac{14}{3}\\ \implies y&=\frac{4}{3}\Big(\frac{8}{3}\Big)-\frac{14}{3}\\ &=\frac{32}{9}-\frac{14}{3}\\ &=\frac{32}{9}-\frac{14\times 3}{3\times 3}\quad\textrm{common denominator}\\ &=\frac{32}{9}-\frac{42}{9}\\ y&=-\frac{10}{9} \end{split} \end{equation}$

So we obtain that the circumcenter is $(\frac{8}{3},-\frac{10}{9})$ , as before.

One of the advantages of having a general formula is that it enables us to see what happens in some special cases.

Example 4

PROVE that if a triangle has two vertices with equal $y$ -coordinates, then the $x$ -coordinate of the circumcenter is the average of the $x$ -coordinates of those two vertices.

To see this, let $y_1=y_2$ in our formula, then the expression for $x$ reduces to:

$\begin{equation*} \begin{split} x&=\frac{x_1^2(y_2-y_3)+x_2^2(y_3-y_1)}{2\Big(x_1(y_2-y_3)+x_2(y_3-y_1)\Big)}\\ &=\frac{x_1^2(y_2-y_3)+x_2^2(y_3-y_2)}{2\Big(x_1(y_2-y_3)+x_2(y_3-y_2)\Big)}\quad\textrm{using}~y_1=y_2\\ &=\frac{x_1^2(y_2-y_3)-x_2^2(y_2-y_3)}{2\Big(x_1(y_2-y_3)-x_2(y_2-y_3)\Big)}\\ &=\frac{(x_1^2-x_2^2)(y_2-y_3)}{2(x_1-x_2)(y_2-y_3)}\\ &=\frac{(x_1+x_2)(x_1-x_2)(y_2-y_3)}{2(x_1-x_2)(y_2-y_3)}\\ &=\frac{x_1+x_2}{2} \end{split} \end{equation}$

Example 5

PROVE that if a triangle has two vertices with equal $x$ -coordinates, then the $y$ -coordinate of the circumcenter is the average of the $y$ -coordinates of those two vertices.

If we let $x_1=x_3$ in our formula, then the expression for $y$ reduces to:

$\begin{equation*} \begin{split} y&=\frac{y_1^2(x_3-x_2)+y_3^2(x_2-x_1)}{2\Big(y_1(x_3-x_2)+y_3(x_2-x_1)\Big)}\\ &=\frac{y_1^2(x_3-x_2)+y_3^2(x_2-x_3)}{2\Big(y_1(x_3-x_2)+y_3(x_2-x_3)\Big)}\quad\textrm{using}~x_1=x_3\\ &=\frac{y_1^2(x_3-x_2)-y_3^2(x_3-x_2)}{2\Big(y_1(x_3-x_2)-y_3(x_3-x_2)\Big)}\\ &=\frac{(y_1^2-y_3^2)(x_3-x_2)}{2(y_1-y_3)((x_3-x_2))}\\ &=\frac{(y_1+y_3)(y_1-y_3)(x_3-x_2)}{2(y_1-y_3)(x_3-x_2)}\\ &=\frac{y_1+y_3}{2} \end{split} \end{equation}$

Since every right triangle can be “oriented” in such a way that its legs are parallel to the $x$ and $y$ axes (making the vertex containing $90^{\circ}$ to have the same $y$ coordinate with one leg and the same $x$ coordinate with the other leg), we see why the circumcenter of a right triangle is at the midpoint of its hypotenuse side. Alternatively, here is a direct proof using “Formula 1” above.

Notation: Preparation for “Formula 2”

Consider $\triangle ABC$ shown below:

We’ll denote the slope of side $AB$ by $m_{c}$ (to mimic the regular notation for side $AB$ ); similarly, the slope of side $AC$ will be denoted by $m_{b}$ . Don’t confuse these with the coordinates of the vertices.

Formula 2

Let $A(a,b),~B(c,d),~C(e,f)$ be the vertices of a triangle $ABC$ . Then the coordinates of the circumcenter are given by:

$\begin{equation*} \begin{split} x&=\frac{(a+c)m_b-(a+e)m_c+(d-f)m_bm_c}{2(m_b-m_c)}\\ y&=\frac{(b+f)m_b-(b+d)m_c+(e-c)}{2(m_b-m_c)} \end{split} \end{equation}$

Let’s consider some examples to illustrate how simple it is to use in practice, so long as the “manual” below is followed:

choose any vertex to start with (call this the “home vertex” — in the diagram above, it is vertex $A$ );
find the slopes of side $AB$ (denoted by $m_c$ ) and side $AC$ (denoted by $m_b$ ) — the two sides that originate from the “home vertex”;
find the numerator of the -coordinate:
- add the $x$ -coordinate of the “home vertex” to the $x$ -coordinate of the vertex on the bottom left, and multiply the sum by the slope of the other side ( $m_b$ )
- add the $x$ -coordinate of the “home vertex” to the $x$ -coordinate of the vertex on the bottom right, and multiply the sum by the slope of the other side ( $m_c$ )
- subtract the $y$ -coordinate of the vertex on the bottom right from the $y$ -coordinate of the vertex on the bottom left, then multiply the difference by $m_b\times m_c$
- combine the three results in the order: $+,-,+$
find the numerator of the -coordinate:
- add the $y$ -coordinate of the “home vertex” to the $y$ -coordinate of the vertex on the bottom right, and multiply the sum by the slope of the same side
- add the $y$ -coordinate of the “home vertex” to the $y$ -coordinate of the vertex on the bottom left, and multiply the sum by the slope of the same side
- subtract the $x$ -coordinate of the vertex on the bottom right from the $x$ -coordinate of the vertex on the bottom left
- combine the three results in the order $+,-,+$
find the denominators of the $x$ and $y$ coordinates (both are $2(m_b-m_c)$ , which is twice the difference of the slopes subtracted in that order);
carry out the divisions;
DONE!!!

Too many steps? Probably confusing? It’s easier than you think.

Example 6

$\triangle ABC$ has vertices at $A(3,2),~B(1,4),~C(5,4)$ . Find the coordinates of its circumcenter.

Let $A(3,2)$ be our “home vertex”. For convenience, we’ve drawn the triangle as follows:

Now, let’s calculate the $x$ -coordinate of the circumcenter:

$\begin{equation*} \begin{split} x&=\frac{(a+c)m_b-(a+e)m_c+(d-f)m_bm_c}{2(m_b-m_c)}\\ &=\frac{(3+1)(1)-(3+5)(-1)+(4-4)(1)(-1)}{2(1--1)}\\ &=\frac{4+8+0}{4}\\ &=\frac{12}{4}\\ &=3 \end{split} \end{equation}$

Next, we find the $y$ -coordinate of the circumcenter:

$\begin{equation*} \begin{split} y&=\frac{(b+f)m_b-(b+d)m_c+(e-c)}{2(m_b-m_c)}\\ &=\frac{(2+4)(1)-(2+4)(-1)+(5-1)}{2(1--1)}\\ &=\frac{6+6+4}{4}\\ &=\frac{16}{4}\\ &=4 \end{split} \end{equation}$

So, the circumcenter is located at $(3,4)$ . Note that if the diagram was drawn accurately, then $\triangle ABC$ is a right triangle with $BC$ as the hypotenuse. The midpoint of $BC$ is $(3,4)$ , the same as the circumcenter obtained. This is an instance of the well-known result we mentioned before.

To show that the answer is independent of which vertex was chosen as the “home vertex”, we’ll solve the example above again, using a different “home vertex”.

Example 7

$\triangle ABC$ has vertices at $A(3,2),~B(1,4),~C(5,4)$ . Find the coordinates of its circumcenter.

Let’s choose $B(1,4)$ as the “home vertex”. As usual, we draw the diagram for convenience:

Ideally we should have $m_a$ instead of $m_b$ in the above diagram, but to avoid confusion, we’ve retained the old notation.

The $x$ -coordinate of the circumcenter is:

$\begin{equation*} \begin{split} x&=\frac{(a+c)m_b-(a+e)m_c+(d-f)m_bm_c}{2(m_b-m_c)}\\ &=\frac{(1+3)(0)-(1+5)(-1)+(2-4)(0)(-1)}{2(0--1)}\\ &=\frac{0+6+0}{2}\\ &=\frac{6}{2}\\ &=3 \end{split} \end{equation}$

Then we find the $y$ -coordinate of the circumcenter:

$\begin{equation*} \begin{split} y&=\frac{(b+f)m_b-(b+d)m_c+(e-c)}{2(m_b-m_c)}\\ &=\frac{(4+4)(0)-(4+2)(-1)+(4-2)}{2(0--1)}\\ &=\frac{0+6+2}{2}\\ &=\frac{8}{2}\\ &=4 \end{split} \end{equation}$

So we obtain $(3,4)$ as before. That should do.

Example 8

$\triangle ABC$ has vertices at $A(-2,-3),~B(0,0),~C(2,1)$ . Find the coordinates of its circumcenter.

This time around we proceed in the traditional way. First, a diagram:

This diagram shows where the circumcenter is ( $Q$ ). Nevertheless, we still carry out the calculations.

The midpoint of $AC$ is $N(0,-1)$ and the slope of $AC$ is $1$ . Thus the slope of the right bisector $NR$ of $AC$ is $-1$ , and its equation is:

(1) $\begin{equation*} y=-x-1 \end{equation*}$

The midpoint of $BC$ is $J(1,0.5)$ and the slope of $BC$ is $\frac{1}{2}$ . Thus the slope of the right bisector $JK$ of $BC$ is $-2$ , and its equation is:

(2) $\begin{equation*} y=-2x+\frac{5}{2} \end{equation*}$

We now have to solve equation(1) and equation(2) simultaneously:

$\begin{equation*} \begin{split} y&=\boxed{-x-1}\\ y&=\boxed{-2x+\frac{5}{2}}\\ -x-1&=-2x+\frac{5}{2}\\ -x+2x&=\frac{5}{2}+1\\ x&=\frac{5}{2}+\frac{2}{2}\\ x&=\frac{7}{2}\\ &\vdots\cdots\vdots\cdots\vdots\cdots\vdots\\ y&=-x-1\\ y&=-\frac{7}{2}-1\\ y&=-\frac{7}{2}-\frac{2}{2}\\ y&=-\frac{9}{2} \end{split} \end{equation}$

Therefore, the circumcenter is located at the point $Q(\frac{7}{2},-\frac{9}{2})$ , as seen in the diagram.

Let’s solve the same question using our formula.

Example 9

$\triangle ABC$ has vertices at $A(-2,-3),~B(0,0),~C(2,1)$ . Find the coordinates of its circumcenter.

We draw a convenient diagram to begin with:

Again we should have used $m_a$ instead of $m_b$ , but we want to stick to the old notation to avoid confusion.

The $x$ -coordinate of the circumcenter is:

$\begin{equation*} \begin{split} x&=\frac{(a+c)m_b-(a+e)m_c+(d-f)m_bm_c}{2(m_b-m_c)}\\ &=\frac{(0+-2)(1/2)-(0+2)(3/2)+(-2-2)(1/2)(3/2)}{2(1/2-3/2)}\\ &=\frac{-2(1/2)-2(3/2)-4(3/4)}{2(-2/2)}\\ &=\frac{-1-3-3}{2(-1)}\\ &=\frac{-7}{-2}\\ &=\frac{7}{2} \end{split} \end{equation}$

The $y$ -coordinate of the circumcenter is:

$\begin{equation*} \begin{split} y&=\frac{(b+f)m_b-(b+d)m_c+(e-c)}{2(m_b-m_c)}\\ &=\frac{(0+1)(1/2)-(0+-3)(3/2)+(2--2)}{2(1/2-3/2)}\\ &=\frac{1/2+9/2+4}{2(-1)}\\ &=\frac{9}{-2}\\ &=-\frac{9}{2} \end{split} \end{equation}$

We obtain the same result $(\frac{7}{2},-\frac{9}{2})$ , as expected.

Example 10

Find the circumcenter of the right triangle $ABC$ with vertices $A(-h,0),~B(h,0),~C(0,h)$ , where $h\neq 0$ .

We begin with a convenient diagram:

The $x$ -coordinate of the circumcenter is:

$\begin{equation*} \begin{split} x&=\frac{(a+c)m_b-(a+e)m_c+(d-f)m_bm_c}{2(m_b-m_c)}\\ &=\frac{(0+-h)(-1)-(0+h)(1)+(0-0)(-1)(1)}{2(-1-1)}\\ &=\frac{h-h+0}{-4}\\ &=\frac{0}{-4}\\ &=0 \end{split} \end{equation}$

The $y$ -coordinate of the circumcenter is:

$\begin{equation*} \begin{split} y&=\frac{(b+f)m_b-(b+d)m_c+(e-c)}{2(m_b-m_c)}\\ &=\frac{(h+0)(-1)-(h+0)(1)+(h--h)}{2(-1-1)}\\ &=\frac{-h-h+2h}{-4}\\ &=\frac{0}{-4}\\ &=0 \end{split} \end{equation}$

So the circumcenter is at $(0,0)$ .

Takeaway

If you’ve calculated the area of a triangle or the volume of a cylinder or the amount of a compound interest, and so on, you’ll agree that there are quite a lot of useful formulas in math — in particular, in geometry.

There are certain occasions where the use of those formulas is inevitable; as such, working from “first principles” (or “re-inventing the wheel”) is inadvisable. On the other hand, there are times when working from “first principles” is the better strategy — not because no formula exists, but because it assists in deepening our understanding of what the formula consists. Knowing which approach to use when you have a choice is a matter of preference, and can be mastered through practice.

Think about this for an analogy. Suppose you’re typing and you need to type or copy paste a particular word which you’ve already typed before, one more time. Will you choose to re-type the word or will you choose to copy paste? It’s a matter of preference definitely. Now suppose that you need to type or copy paste a particular paragraph which has already appeared before, one more time. Will you choose to re-type the paragraph or will you just copy paste?

Tasks

Find the circumcenter of an equilateral triangle whose vertices are $A(-a,0),~B(a,0),~C(0,a\sqrt{3});\quad a\neq 0$ .
Find the circumcenter of the isosceles triangle whose vertices are located at $P(-a,0),~Q(a,0),~R(0,a);\quad a\neq 0$ .
Let $(x_1,y_1),~(x_2,y_2),~(x_3,y_3)$ be the vertices of a triangle. If
$\begin{equation*} \begin{split} (x_3-x_1)^2+(y_3-y_1)^2&= (x_2-x_1)^2+(y_2-y_1)^2\\ &+(x_3-x_2)^2+(y_3-y_2)^2,\end{split} \end{equation}$

PROVE that:

$\frac{y_3-y_2}{x_3-x_2}\times\frac{y_2-y_1}{x_2-x_1}=-1.$

[The converse is also true. Note that the result holds whether or not we’re dealing with the vertices of a triangle. However, when applied to triangles, it tells us that we can either use lengths or slopes (not both) to verify that a triangle contains $90^{\circ}$ .]
PROVE that:
$\begin{equation*} \begin{split} x_1^2(y_2-y_3)+x_2^2(y_3-y_1)+x_3^2(y_1-y_2)&=y_1(x_3^2-x_2^2)+\\ &y_2(x_1^2-x_3^2)+y_3(x_2^2-x_1^2) \end{split} \end{equation}$

[It follows that we have two expressions for the numerator of the $x$ -coordinate of our circumcenter formula 1. Same thing applies to the $y$ -coordinate.]
PROVE that:
$\begin{equation*} \begin{split} x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)&=y_1(x_3-x_2)+\\ &y_2(x_1-x_3)+y_3(x_2-x_1) \end{split} \end{equation}$

[It follows that we have two expressions for the denominator of the $x$ -coordinate of our circumcenter formula 1. Same thing applies to the $y$ -coordinate.]