Prove that the centres of the three circles x2 + y2 – 4x – 6y – 12 = 0, x2 + y2 + 2x + 4y – 10 = 0 and x2 + y2 – 10x – 16y – 1 = 0 are collinear.
Given circles are:
(i) x2 + y2 - 4x - 6y - 12 = 0
(ii) x2 + y2 + 2x + 4y - 10 = 0
(iii) x2 + y2 - 10x - 16y - 1 = 0
We know that for a circle x2 + y2 + 2ax + 2by + c = 0 - - (1)
⇒ Centre = (- a, - b)
⇒ Radius = 
Comparing (i) with (1) we get,
⇒ Centre(A) = 
⇒ A = (2,3)
Comparing (ii) with (1) we get,
⇒ Centre(B) = 
⇒ B = (- 1, - 2)
Comparing (iii) with (1) we get,
⇒ Centre(C) = 
⇒ C = (5,8)
We need to show that A, B, and C are collinear.
We find the line passing through any two points and check whether the third point is present on it or not, by substituting the point in the line.
Let us assume the line passes through the points A, B.
We know that the equation of the line passing through the points (x1,y1) and (x2,y2) is 
.
⇒ 
⇒ 
⇒ 3(y - 3) = 5(x - 2)
⇒ 3y - 9 = 5x - 10
⇒ 5x - 3y - 1 = 0 ..... (2)
Substituting point C(5,8) in (2), we get
⇒ 5(5) - 3(8) - 1
⇒ 25 - 24 - 1
⇒ 0
∴ The centres of the circle are collinear.