Show that the circles $\;$ $x^2 + y^2 - 2x + 6y + 6 = 0$ $\;$ and $\;$ $x^2 + y^2 - 5x + 6y + 15 = 0$ $\;$ touch each other.
$\begin{aligned}
\text{Given equations of circles: } & x^2 + y^2 - 2x + 6y + 6 = 0 \;\;\; \cdots \; (1) \\\\
& x^2 + y^2 - 5x + 6y + 15 = 0 \;\;\; \cdots \; (2) \\\\
\text{Standard equation of circle: } & x^2 + y^2 + 2gx + 2fy + c = 0 \;\;\; \cdots \; (3)
\end{aligned}$
Comparing equation $(1)$ with equation $(3)$ gives
$2 g_1 = -2 \implies g_1 = -1$; $\;\;$ $2f_1 = 6 \implies f_1 = 3$; $\;\;$ $c_1 = 6$
$\therefore$ $\;$ Center of equation $(1)$ is $= C_1 = \left(-g_1, -f_1\right) = \left(1, -3\right)$
Radius of equation $(1)$ is $= r_1 = \sqrt{g_1^2 + f_1^2 - c}$
i.e. $r_1 = \sqrt{\left(-1\right)^2 + \left(3\right)^2 - 6} = \sqrt{1 + 9 - 6} = 2$
Comparing equation $(2)$ with equation $(3)$ gives
$2g_2 = - 5 \implies g_2 = \dfrac{-5}{2}$; $\;\;$ $2 f_2 = 6 \implies f_2 = 3$; $\;\;$ $c_2 = 15$
$\therefore$ $\;$ Center of equation $(2)$ is $= C_2 = \left(-g_2, -f_2\right) = \left(\dfrac{5}{2}, -3\right)$
Radius of equation $(2)$ is $= r_2 = \sqrt{g_2^2 + f_2^2 - c_2}$
i.e. $r_2 = \sqrt{\left(\dfrac{-5}{2}\right)^2 + \left(3\right)^2 - 15} = \sqrt{\dfrac{25}{4} + 9 - 15} = \dfrac{1}{2}$
Now, $C_1 C_2 = \sqrt{\left(\dfrac{5}{2} - 1\right)^2 + \left(-3 + 3\right)^2} = \sqrt{\dfrac{9}{4}} = \dfrac{3}{2}$
and $r_1 - r_2 = 2 - \dfrac{1}{2} = \dfrac{3}{2}$
$\therefore$ $\;$ $C_1 C_2 = r_1 - r_2$
$\implies$ The circles given by equations $(1)$ and $(2)$ touch internally.