circumcenter = (-1,0)
The circumcenter of a triangle is the intersection of the perpendicular bisectors of the sides of the triangle. So let's calculate a couple of the bisectors and determine their intersection.
Slope AB = (3 - -3)/(2 - -4) = (3+3)/(2+4) = 6/6 = 1
Perpendicular bisector will have a slope of -1 and will pass through point ((2-4)/2,(3-3)/2) = -2/2,0/2) = (-1,0)
Equation is of the form
y = -x + b
Substitute known point
0 = -(-1) + b
0 = 1 + b
-1 = b
So the equation for the perpendicular bisector of AB is
y = -x - 1
Now let's calculate the perpendicular bisector of BC
Slope BC = (-3 - -3)/(-4 - 2) = (-3 + 3) / (-6) = 0/-6 = 0. This means that the
line is horizontal and that the perpendicular bisector will be a vertical line with infinite slope. A point that line will pass through is ((-4 + 2)/2, (-3 + -3)/2) =
(-2/2, 0/2) = (-1,0). So the equation for the line is:
x = -1
Now we want the intersection between
x = -1 and y = -x - 1
Since we know that x has to be -1, just substitute it into the 2nd equation.
y = -x - 1
y = -(-1) - 1
y = 1 - 1
y = 0
So the circumcenter is (-1,0).
Let's verify that. The distance from the circumcenter to each vertex of the triangle will be the same. Using the Pythagorean theorem, C^2 = A^2 + B^2. We're not going to bother taking the square root since if the squares are equal, then square roots will also be equal.
Distance^2 from (2,3):
(2- -1)^2 + (3-0)^2 = 3^2 + 3^2 = 9 + 9 = 18
Distance^2 from (-4,-3):
(-4 - -1)^2 + (-3 - 0)^2 = -3^2 + -3^2 = 9 + 9 = 18
Distance^2 from (2,-3):
(2 - -1)^2 + (-3 - 0)^2 = 3^2 + -3^2 = 9 + 9 = 18
The distances to all three vertexes is identical, so (-1,0) is verified as the circumcenter.
