The coordinates of the vertices of a triangle are (1, 2),(2, 3), (3, 1). Find the coordinates of the centre of its circumcircle and the circumradius.

Let the centre of the circumcircle be (x,y)

Since, centre of circumcircle is a point which is equidistant from all the points,

(x – 1)² + (y – 2)² = (x – 2)² + (y – 3)² = (x – 3)² + (y – 1)²

⇒ x² + y² – 2x – 4y + 5 = x² + y² – 4x – 6y + 13 = x² + y² – 6x – 2y + 10

⇒ – 2x – 4y + 5 = – 4x – 6y + 13 = – 6x – 2y + 10

– 2x – 4y + 5 = – 4x – 6y + 13

⇒ 2x + 2y = 8

⇒ x + y = 4 …….(1)

– 4x – 6y + 13 = – 6x – 2y + 10

⇒ 2x – 4y = – 3…..(2)

Substituting x = 4 – y from eq(1)

2(4 – y) – 4y = – 3

⇒ 8 – 2y – 4y = – 3

⇒ 8 – 6y = – 3

⇒ – 6y = – 11

⇒

And

x = 4 – y

⇒

Hence coordinates of the centre of the circumcircle of the triangle = )

And its radius =

=

= units.