We use x1+x2+x3/3, y1+y2+y3/3 to find the centroid of a triangle. Can I do the same to find circumcentre?

Answer 1

No. The centroid and circumcenter (and incenter) are (likely) different points of concurrency.

Finding the circumcenter is a bit longer process: We find equations of the perpendicular bisectors of two sides, then find the point of intersection of those two perpendicular bisectors. You may choose any pair of triangle sides and find the perpendicular bisectors in order to find the circumcenter.

Here's an example:

Given A: (-3 , 1) , B: (5 , 5) , and C: (2 , -4) find the circumcenter of ΔABC.

(Note: M_AB denotes the midpoint of segment AB, while m_AB denotes the slope of segment AB.)

Finding the equation of the ⊥ bisector of AB: M_AB = ((-3 + 5)/2 , (1 + 5)/2) = (1 , 3) ; m_AB = (5 - 1)/(5 - (-3)) = 1/2. The eqn of the ⊥ bisector of AB is y - 3 = -2(x - 1) or y = -2x + 5.

Finding the equation of the ⊥ bisector of BC: M_BC = ((2 + 5)/2 , (-4 + 5)/2) = (7/2 , 1/2) ; m_BC = (5 - (-4))/(5 - 2) = 3. The eqn of the ⊥ bisector of AB is y - 1/2 = -1/3(x - 7/2) or y = -1/3x + 5/3.

The pt. of intersection between these two lines can be found by setting both righthand sides = to each other:

-2x + 5 = -1/3x + 5/3

-5/3x = -10/3

x = 2 ; y = -2(2) + 5 = 1. The circumcenter is (2 , 1).

Demonstrating that (2,1) is equidistant from A and B and C is left as an exercise for the reader.