Find the common tangents of the circles $\;$ $x^2 + y^2 - 3x - 4y = 0$, $\;$ $x^2 + y^2 - 21x + 90 = 0$.
The given circles are
$x^2 + y^2 - 3x - 4y = 0$ $\;\;\; \cdots \; (1)$
$x^2 + y^2 - 21x + 90 = 0$ $\;\;\; \cdots \; (2)$
Comparing equations $(1)$ and $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
for circle $(1)$: $\;$ $g_1 = \dfrac{-3}{2}, \;\; f_1 = -2, \;\; c_1 = 0$
center of circle $= C_1 = \left(-g_1, -f_1\right) = \left(\dfrac{3}{2}, 2\right)$,
radius of circle $= r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{\dfrac{9}{4} + 4 - 0} = \dfrac{5}{2}$
for circle $(2)$: $\;$ $g_2 = \dfrac{-21}{2}, \;\; f_2 = 0, \;\; c_2 = 90$
center of circle $= C_2 = \left(-g_2, -f_2\right) = \left(\dfrac{21}{2}, 0\right)$
radius of circle $= r_2 = \sqrt{g_2^2 + f_2^2 - c_2} = \sqrt{\dfrac{441}{4} + 0 - 90} = \dfrac{9}{2}$
Ratio $\;$ $\dfrac{r_1}{r_2} = \dfrac{5}{9}$
Let $S_1$ be the external center of similitude of circles $(1)$ and $(2)$.
Then, the point $S_1$ divides the line joining $C_1C_2$ externally in the ratio $\dfrac{r_1}{r_2}$.
$\therefore \;$ $S_1 = \left(\dfrac{5 \times \dfrac{21}{2} - 9 \times \dfrac{3}{2}}{5-9}, \dfrac{5 \times 0 - 9 \times 2}{5-9}\right) = \left(\dfrac{-39}{4}, \dfrac{9}{2}\right)$
Let $S_2$ be the internal center of similitude of circles $(1)$ and $(2)$.
Then, the point $S_2$ divides the line joining $C_1C_2$ internally in the ratio $\dfrac{r_1}{r_2}$.
$\therefore \;$ $S_2 = \left(\dfrac{5 \times \dfrac{21}{2} + 9 \times \dfrac{3}{2}}{5+9}, \dfrac{5 \times 0 + 9 \times 2}{5+9}\right) = \left(\dfrac{33}{7}, \dfrac{9}{7}\right)$
Let the equation to either of the tangents passing through $S_1$ be
$y - \dfrac{9}{2} = m \left(x + \dfrac{39}{4}\right)$
i.e. $\;$ $mx - y + \dfrac{39}{4}m + \dfrac{9}{2} = 0$ $\;\;\; \cdots \; (3)$
Equation $(3)$ will touch circle $(1)$ if perpendicular distance from $C_1$ to equation $(3)$ is equal to $r_1$.
i.e. $\;$ $\dfrac{\dfrac{3}{2}m - 2 + \dfrac{39}{4}m + \dfrac{9}{2}}{\sqrt{m^2 + 1}} = \pm \dfrac{5}{2}$
i.e. $\;$ $\dfrac{45m + 10}{4 \sqrt{m^2 + 1}} = \pm \dfrac{5}{2}$
i.e. $\;$ $45m + 10 = \pm 10 \sqrt{m^2 + 1}$
i.e. $\;$ $2025 m^2 + 900 m + 100 = 100 + 100 m^2$
i.e. $\;$ $1925 m^2 + 900 m = 0$ $\;\;\; \cdots \; (4)$
Solving equation $(4)$ we get $\;\;$ $m = 0$, $\;$ or $\;$ $m = \dfrac{-36}{77}$
Substituting the value of $m$ in equation $(3)$ gives the required tangents.
$\therefore \;$ The required tangents are
$-y + \dfrac{9}{2} = 0$; $\;\;\;$ $\dfrac{-36}{77}x - y + \dfrac{39}{4} \times \left(\dfrac{-36}{77}\right) + \dfrac{9}{2} = 0$
i.e. $\;$ $2y - 9 = 0$ $\;\;\; \cdots \; (5a)$; $\;\;\;$ $36 x + 77y + \dfrac{9}{2} = 0$ $\;\;\; \cdots \; (5b)$
Let the equation to either of the tangents passing through $S_2$ be
$y - \dfrac{9}{7} = m_1 \left(x - \dfrac{33}{7}\right)$
i.e. $\;$ $m_1 x - y + \dfrac{9}{7} - \dfrac{33}{7} m = 0$ $\;\;\; \cdots \; (6)$
Equation $(6)$ will touch circle $(1)$ if perpendicular distance from $C_1$ to equation $(6)$ is equal to $r_1$.
i.e. $\;$ $\dfrac{\dfrac{3}{2} m_1 - 2 + \dfrac{9}{7} - \dfrac{33}{7} m_1}{\sqrt{m_1^2 + 1}} = \pm \dfrac{5}{2}$
i.e. $\;$ $\dfrac{-45 m_1 - 10}{14\sqrt{m_1^2 + 1}} = \pm \dfrac{5}{2}$
i.e. $\;$ $-45m_1 - 10 = \pm 35 \sqrt{m_1^2 + 1}$
i.e. $\;$ $2025 m_1^2 + 900 m_1 + 100 = 1225 + 1225 m_1^2$
i.e. $\;$ $32m_1^2 + 36m_1 - 45 = 0$ $\;\;\; \cdots \; (7)$
Solving equation $(7)$ we get $\;\;$ $m_1 = \dfrac{-15}{8}$, $\;$ or $\;$ $m_1 = \dfrac{3}{4}$
Substituting the value of $m_1$ in equation $(6)$ gives the required tangents.
$\therefore \;$ The required tangents are
$\dfrac{-15}{8} x - y + \dfrac{9}{7} - \dfrac{33}{7} \times \left(\dfrac{-15}{8}\right) = 0$; $\;\;\;$ $\dfrac{3}{4} x - y + \dfrac{9}{7} - \dfrac{33}{7} \times \dfrac{3}{4} = 0$
i.e. $\;$ $15x + 8y - 81 = 0$ $\;\;\; \cdots \; (8a)$; $\;\;\;$ $3x - 4y - 9 = 0$ $\;\;\; \cdots \; (8b)$
Equations $(5a)$, $(5b)$, $(8a)$ and $(8b)$ are the required common tangents to circles given by equations $(1)$ and $(2)$.