Let A(-3,1), B(0,-2) and C(1,3) be the vertices of the triangle.

Let O(h,k) be the circumcentre of the triangle.

OA , OB and OC are the radii of the circle.

OA = OB………………….(i)

By distance formula, OA = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

√[(h–3)^{2}+(k-1)^{2}] = √[(h+3)^{2}+(k-1)^{2}] ……..(ii)

OB = √[(h-0)^{2}+(k-(-2))^{2}]

= √[(h^{2}+(k+2)^{2}] …………………………..(iii)

Equating (ii) and (iii)

√[(h+3)^{2}+(k-1)^{2}] = √[(h^{2}+(k+2)^{2}]

Squaring both sides[(h+3)^{2}+(k-1)^{2}] = [(h^{2}+(k+2)^{2}]

h^{2}+6h+9+k^{2}-2k+1 = h^{2}+k^{2}+4k+4

6h-2k+10 = 4k+4

6h-6k = -6

h-k = -1 ……….(a) [Dividing both sides by 6]

OB = OC [radii of same circle]

OC = √[(h-1)^{2}+(k-3)^{2}]

= √[h^{2}-2h+1+k^{2}-6k+9]

=√[ h^{2}-2h+k^{2}-6k+10] ……..(iv)

Equating (iii) and (iv)

√[(h^{2}+(k+2)^{2}] = √[ h^{2}-2h+k^{2}-6k+10]

Squaring both sides

(h^{2}+(k+2)^{2} = h^{2}-2h+k^{2}-6k+10

(h^{2}+k^{2}+4k+4 = h^{2}-2h+k^{2}-6k+10

4k+4 = -2h-6k +10

4k+6k+2h = 10-4

10k+2h = 6

Divide by 2 on both sides , we get

5k+h = 3 ………(b)

Solving (a) and (b)

h-k = -1…….(a)

h+5k = 3 ……..(b)

(a)-(b)

h-k = -1

-h-5k = -3

-6k = -4

k = -4/-6 = 2/3

Substitute the value of k in (a)

h-k = -1

h-(2/3) = -1

h = -1+(2/3)

h= (-3/3)+(2/3)

h = -1/3

Hence the co-ordinates of circumcentre of triangle are (-1/3 , 2/3)