Circumcentre of a Triangle from Coordinates: Worked Example

Grades 9-10

Question: The vertices of a triangle are A(0,0), B(8,0), and C(0,6). Find the coordinates of its circumcenter

To find the coordinates of the circumcenter of

\triangle ABC

with vertices

A(0, 0)

B(8, 0)

, and

C(0, 6)

, we can use properties of coordinate geometry and circle theorems.

First, let's identify the nature of the triangle.

Since the x-axis is perpendicular to the y-axis, we have:

\angle BAC = 90^\circ

Therefore,

\triangle ABC

is a right-angled triangle at

A

In a right-angled triangle, the circumcenter has a very specific and useful property:

The circumcenter is the midpoint of the hypotenuse [ $\angle$ in semi-circle].
The hypotenuse is the side opposite the right angle ( $90^\circ$ ), which in this case is the line segment $BC$ .

Let

P(x, y)

be the circumcenter of

\triangle ABC

. Since

P

is the midpoint of

BC

\begin{align*} x &= \frac{x_B + x_C}{2} \\ x &= \frac{8 + 0}{2} = 4 \\ y &= \frac{y_B + y_C}{2} \\ y &= \frac{0 + 6}{2} = 3 \end{align*}

Thus, the coordinates of the circumcenter are

(4, 3)

The circumcenter is the intersection of the perpendicular bisectors of the sides.

The midpoint of $AB$ is $(4, 0)$ . Since $AB$ is horizontal, its perpendicular bisector is the vertical line $x = 4$ .
The midpoint of $AC$ is $(0, 3)$ . Since $AC$ is vertical, its perpendicular bisector is the horizontal line $y = 3$ .
The intersection of $x = 4$ and $y = 3$ is $(4, 3)$ .

The coordinates of the circumcenter are

(4, 3)

. The radius of the circumcircle (the circumradius) is the distance from

(4, 3)

to any vertex, which is

\sqrt{4^2 + 3^2} = 5

units.