prove that the centres of the three circles x^2+y^2-4x-6y-12=0, x^2+y^2+2x+4y-5=0 and x^2+y^2-10x-16y+7=0 are collinear

First take the first equation :-

x² + y² -4x -6y +12 =0
We can split 12 as 4+9-1.

Therefore the equation becomes - x² -4x +4 +y² -6y +9 -1 =0 (This splitting on constant is only done to form whole squares)

therefore, the equation becomes (x-2)² + (y-3)² =1
Therefore, its centre is (2,3)

Similarly, eq 2 i.e. x² +y² +2x +4y -5=0 will be equal to x² +2x +1 +y2 +4y +4 -10 =0
It will be equal to (x+1)² + (y+2)² =10

Centre = (-1,-2)

In the same way, eq 3 will be equal to x² -10x +25 +y² -16y +64 -82 =0

which will be equal to (x-5)² +(y-8)² =82
Centre = (5,8)

If you take the slope of any of the two points of (2,3) , (-1,-2) and (5,8) it will be equal to 5/3.
Therefore the points are collinear.

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