Find the coordinates of the circumcentre of a triangle whose vertices are (-3,1), (0,-2) and (1,3).

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 = √[(x₂-x₁)²+(y₂-y₁)²]
√[(h–3)²+(k-1)²] = √[(h+3)²+(k-1)²] ……..(ii)
OB = √[(h-0)²+(k-(-2))²]
= √[(h²+(k+2)²] …………………………..(iii)
Equating (ii) and (iii)
√[(h+3)²+(k-1)²] = √[(h²+(k+2)²]
Squaring both sides[(h+3)²+(k-1)²] = [(h²+(k+2)²]
h²+6h+9+k²-2k+1 = h²+k²+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]
= √[h²-2h+1+k²-6k+9]
=√[ h²-2h+k²-6k+10] ……..(iv)
Equating (iii) and (iv)
√[(h²+(k+2)²] = √[ h²-2h+k²-6k+10]
Squaring both sides
(h²+(k+2)² = h²-2h+k²-6k+10
(h²+k²+4k+4 = h²-2h+k²-6k+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)