# Three points on a circle are o{0,0} , A{-2,1} , B{ -3,2}. Find the center of the circle

Given three points O(0, 0), A(-2, 1) and B(-3, 2)

Let the centre of the circle be C(a, b)

Hence, OC = AC = BC (as they will be the radius of the circle)

So, OC^2 = AC^2 = BC^2

(a – 0)^2 + (b – 0)^2 = (a + 2)^2 + (b – 1)^2 = (a + 3)^2 + (b – 2)^2

a^2 + b^2 = a^2 + 4a + 4 + b^2 – 2b + 1 = a^2 + 6a + 9 + b^2 – 4b + 4

On equating and solving first two terms, we get,

a^2 + b^2 = a^2 + 4a + 4 + b^2 – 2b + 1

4a – 2b + 5 = 0 …..(i)

On equating and solving second & third terms, we get,

a^2 + 4a + 4 + b^2 – 2b + 1 = a^2 + 6a + 9 + b^2 – 4b + 4

2a – 2b + 8 = 0 …..(ii)

Solving eq(i) & (ii), we get,

a = 3/2 and b = 11/2

Hence, centre of circle passing through O{0,0} , A{-2,1} , B{-3,2} is (3/2, 11/2)