Prove that the centers of the three circles $x^2 + y^2 = 1$, $\;$ $x^2 + y^2 + 6x - 2y = 1$ $\;$ and $\;$ $x^2 + y^2 - 12x + 4y = 1$ are collinear.
Equations of the circles are:
$x^2 + y^2 = 1$ $\;\;\; \cdots \; (1)$
$x^2 + y^2 + 6x - 2y = 1$ $\;\;\; \cdots \; (2)$
$x^2 + y^2 - 12x + 4y = 1$ $\;\;\; \cdots \; (3)$
Center of circle given by equation $(1)$ is $C_1 = \left(0,0\right)$
The standard equation of circle is: $\hspace{1em}$ $x^2 + y^2 + 2gx + 2 fy + c = 0$ $\;\;\; \cdots \; (4)$
Comparing equation of the circle $(2)$ with the standard equation $(4)$ gives
$2g = 6 \implies g = 3$; $\;\;\;$ $2f = -2 \implies f = -1$
$\therefore$ $\;$ Center of circle given by equation $(2)$ is $C_2 = \left(-g , -f\right) = \left(-3, 1\right)$
Comparing equation of the circle $(3)$ with the standard equation $(4)$ gives
$2g = -12 \implies g = -6$; $\;\;\;$ $2f = 4 \implies f = 2$
$\therefore$ $\;$ Center of circle given by equation $(3)$ is $C_3 = \left(-g , -f\right) = \left(6, -2\right)$
Now, slope of $C_1 C_2 = m_1 = \dfrac{1-0}{-3-0} = \dfrac{-1}{3}$
Slope of $C_2 C_3 = m_2 = \dfrac{- 2 - 1}{6 + 3} = \dfrac{-3}{9} = \dfrac{-1}{3}$
Slope of $C_3 C_1 = m_3 = \dfrac{-2 - 0}{6 - 0} = \dfrac{-2}{6} = \dfrac{-1}{3}$
$\because$ $\;$ $m_1 = m_2 = m_3$ $\implies$ $C_1$, $C_2$, $C_3$ are collinear.