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.

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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.

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.

Answer:

Given,

x² + y² – 4x – 6y – 12 = 0

Centre ( – g₁, – f₁) = (2, 3)

x² + y² + 2x + 4y – 5 = 0

Centre ( – g₂, – f₂) = ( – 1, – 2)

x² + y² – 10x – 16y + 7 = 0

Centre ( – g₃, – f₃) = (5, 8)