Question

Find the coordinates of the circumcentre of the triangle whose vertices are (8,6) , (8,-2) and (2,-2) also, find its circumradius.

Answer 1

Let the coordinates of the circumcentre of the triangle be (x, y). Circumcentre of a triangle is equidistant from each of the vertices. Distance between (8, 6) and (x, y) = Distance between (8, -2) and (x, y)

√[(x - 8)^2+ (y - 6)^2] = √[(x - 8)^2+ (y + 2)^2]

[(x - 8)^2 + (y - 6)^2] = [(x - 8)^2 + (y + 2)^2]

(y - 6)^2= (y + 2)^2

y^2 + 36 - 12y = y^2 + 4y + 4

36 - 12y = 4y + 4

16y = 32
y = 2

Distance between (2, -2) and (x, y) = Distance between (8, -2) and (x, y)

√[(x - 2)^2 + (y + 2)^2] = √[(x - 8)^2 + (y + 2)^2]

[(x - 2)^2 + (y + 2)^2] = [(x - 8)^2 + (y + 2)^2]

(x - 2)^2 = (x - 8)^2

x^2 + 4 - 4x = x^2- 16x + 64

4 - 4x = -16x + 64

12x = 60

x = 5.

Hence, the coordiantes of the circumcentre of the triangle are (5, 2).

Circumradius = √[(5 - 8)^2+ (2 - 6)^2] = √(9 + 16) = √25 = 5 units.