Question

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.

Answer 1

The given equations of the circles are as follows:
x² + y² − 4x − 6y − 12 = 0,   ...(1)
x² + y² + 2x + 4y − 10 = 0    ...(2)
And, x² + y² − 10x − 16y − 1 = 0    ...(3)

The centre of circle (1) is (2, 3).
The centre of circle (2) is (−1, −2).
The centre of circle (3) is (5, 8).

The area of the triangle formed by the points (2, 3), (−1, −2) and (5, 8) is \[\frac{1}{2}\left| 2\left( - 10 \right) - 1\left( 5 \right) + 5\left( 5 \right) \right| = \frac{1}{2}\left| - 25 + 25 \right| = 0\]

Hence, the centres of the circles x² + y² − 4x − 6y − 12 = 0, x² + y² + 2x + 4y − 10 = 0 and x² + y² − 10x − 16y − 1 = 0 are collinear.