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}

⇒ x^{2} + y^{2} – 2x – 4y + 5 = x^{2} + y^{2} – 4x – 6y + 13 = x^{2} + y^{2} – 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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