Answer :

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.


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