Find the circumcenter of the triangle (-4,0),(0,-8),(0,0)

Answer 1

circumcenter = (x,y) where distance is the same from (x,y) to each vertex

from (x,y) to (-4,0), (0,-8) and (0,0)

it's a right triangle in quadrant III

use the Pythagorean Theorem on each vertex to (x,y), d= the distance

d^2 = (y-0)^2 + (x+4)^2 = (y+8)^2 + (x-0)^2 = x^2 + y^2

y^2 + x^2+8x +16 = d^2

y^2 + x^2 = d^2

subtract to get

8x +16 = 0

x=-2

y^2 +16y + 64 +x^2 = d^2

y^2 +x^2 = d^2

subtract to get

16y +64 = 0

y=-4

circumcenter = (-2,-4)

d = sqr(16+4) = sqr20 = 2sqr5 = distance from (-2,-4) to each vertex