The Perpendicular Bisectors and the Circumcenter of a Triangle

The perpendicular bisectors of the sides of a triangle are concurrent at a point called circumcenter.

This point of concurrency is the center of the circle circumscribing the triangle and is equidistant from the vertices of the triangle.

This post is the second to the last part of a series of posts about the points of concurrency of a triangle's parts.

Here's a list of my previous posts about concurrent parts of a triangle:

• orthocenter: the point of concurrency of a triangle's altitudes (see Altitudes and Orthocenter of a Triangle)
• centroid: the point of concurrency of a triangle's medians (see Medians And Centroid Of A Triangle)
• incenter: the point of concurrency of a triangle's angle bisectors (see The Angle Bisectors and the Incenter of a Triangle)

The Problem

Given the vertices A(-5, 2), B(2, 3) and C(-1, -6) of triangle ABC, find the point of concurrency of the perpendicular bisectors of its sides. Find the equation of the circle circumscribed about the triangle.

Discussion

Referring to the given illustration above, line segments OP, OQ and OR are the perpendicular bisectors of the sides of scalene triangle ABC.

Their point of concurrency, point O, is called circumcenter because it is the center of the circle circumscribing the triangle.

Line segments AO, BO and CO are the radii of circle O.

The problem illustrates the fact that any triangle can always be circumscribed by a circle.

In other words, a set of three noncollinear points which are on the same circle (concyclic) can always be found.

Solution

Coordinates of Circumcenter: (-1, -1)



Midpoints of the Sides

• side AB: P(- 3/2, 5/2)
• side BC: Q(1/2, - 3/2)
• side AC: R(-3, -2)

Distance Between Two Points

• distance_AB = √ 50
• distance_BC = √ 90
• distance_AC = √ 80
• distance_AO = distance_BO = distance_CO = √ 25
• distance_AP = distance_BP = √ 25/2
• distance_BQ = distance_CQ = √ 45/2
• distance_AR = distance_CR = √ 20

Equations of the Line Segments

• side AB: x - 7y + 19 = 0; x-intercept = -19; y-intercept = 19/7
• side BC: 3x - y - 3 = 0; x-intercept = 1; y-intercept = -3
• side AC: 2x + y + 8 = 0; x-intercept = - 4; y-intercept = - 8
• perpendicular bisector OP: 7x + y + 8 = 0; x-intercept = - 8/7; y-intercept = - 8
• perpendicular bisector OQ: x + 3y + 4 = 0; x-intercept = - 4; y-intercept = - 4/3
• perpendicular bisector OR: x - 2y - 1 = 0; x-intercept = 1; y-intercept = - 1/2
• radius AO: 3x + 4y + 7 = 0; x-intercept = - 7/3; y-intercept = - 7/4
• radius BO: 4x - 3y + 1 = 0; x-intercept = - 1/4; y-intercept = 1/3
• radius CO: x = - 1; x-intercept = - 1

Equation of the Circle:

( x + 1 )² + ( y + 1 )² = (√ 25)²

x² + y² + 2x + 2y - 23 = 0

Slopes of the Line Segments

• m_AB = 1/7
• m_BC = 3
• m_AC = - 2
• m_AO = - 3/4
• m_BO = 4/3
• m_OP = - 7
• m_OQ = - 1/3
• m_OR = 1/2

Interior Angles of the Triangle

• angle A = 71.6°
• angle B = 63.4°
• angle C = 45°