Show that the circles $\;$ $x^2 + y^2 -8x - 4y + 10 = 0$ $\;$ and $\;$ $x^2 + y^2 - 10x - 10 y + 10 = 0$ $\;$ touch each other at the point $\left(3, -1\right)$. Also find the equation of the common tangent.
Equations of the given circles are
$x^2 + y^2 - 8x - 4y + 10 = 0$ $\;\;\; \cdots \; (1)$
$x^2 + y^2 - 10x - 10y + 10 = 0$ $\;\;\; \cdots \; (2)$
Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2 f_1 y + c_1 = 0$ $\;$ gives
$g_1 = -4, \; f_1 = -2, \; c_1 = 10$
Center of circle [equation $(1)$] is $\;$ $C_1 = \left(-g_1, -f_1\right) = \left(4, 2\right)$
Radius of circle [equation $(1)$] is
$r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{16 + 4 - 10} = \sqrt{10}$
Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2 f_2 y + c_2 = 0$ $\;$ gives
$g_2 = -5, \; f_2 = -5, \; c_2 = 10$
Center of circle [equation $(2)$] is $\;$ $C_2 = \left(-g_2, -f_2\right) = \left(5, 5\right)$
Radius of circle [equation $(2)$] is
$r_2 = \sqrt{g_2^2 + f_2^2 - c_2} = \sqrt{25 + 25 - 10} = \sqrt{40} = 2 \sqrt{10}$
Distance between the centers of the two circles
$= C_1 C_2 = \sqrt{\left(4 - 5\right)^2 + \left(2 - 5\right)^2} = \sqrt{1 + 9} = \sqrt{10}$
$\left|r_1 - r_2\right| = \left|\sqrt{10} - 2 \sqrt{10}\right| = \sqrt{10}$
i.e. $\;$ $C_1 C_2 = \left|r_1 - r_2\right|$
$\implies$ The two circles touch each other internally.
Subtracting equations $(1)$ and $(2)$ gives
$2x + 6y = 0$ $\implies$ $x = - 3y$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$ equation $(1)$ becomes
$\left(-3y\right)^2 + y^2 - 8 \times \left(-3y\right) - 4y + 10 = 0$
i.e. $\;$ $9 y^2 + y^2 + 24 y - 4y + 10 = 0$
i.e. $\;$ $10 y^2 + 20 y + 10 = 0$
i.e. $\;$ $y^2 + 2y + 1 = 0$
i.e. $\;$ $\left(y + 1\right)^2 = 0$
$\implies$ $y = -1$
Substituting the value of $y$ in equation $(3)$ gives
$x = - 3 \times \left(-1\right) = 3$
$\therefore \;$ The point of contact of the two circles is $\left(x_1, y_1\right) = \left(3, -1\right)$
Equation of common tangent at the point $\left(x_1, y_1\right)$ to circle given by equation $(1)$ is
$xx_1 + yy_1 + g_1 \left(x + x_1\right) + f_1 \left(y + y_1\right) + c_1 = 0$
i.e. $\;$ $3x - y - 4 \left(x + 3\right) - 2 \left(y - 1\right) + 10 = 0$
i.e. $\;$ $3x - y - 12 - 2y + 2 + 10 = 0$
i.e. $\;$ $x + 3y = 0$