- Ta có :

\(\left( {{C_1}} \right):{x^2} + {\left( {y - 2} \right)^2} = 9 \Rightarrow {I_1}\left( {0;2} \right),{R_1} = 3,\quad \left( {{C_2}} \right):{\left( {x - 3} \right)^2} + {\left( {y + 4} \right)^2} = 9 \Rightarrow {I_2}\left( {3; - 4} \right),{R_2} = 3\)

- Nhận xét : \({I_1}{I_2} = \sqrt {9 + 4}  = \sqrt {13}  < 3 + 3 = 6 \Rightarrow \left( {{C_1}} \right)\) không cắt \((C_2)\)

- Gọi \(d:ax + by + c = 0\) (\({a^2} + {b^2} \ne 0\)) là tiếp tuyến chung , thế thì: \(d\left( {{I_1},d} \right) = {R_1},d\left( {{I_2},d} \right) = {R_2}\)

\(\begin{array}{l}
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{{\left| {2b + c} \right|}}{{\sqrt {{a^2} + {b^2}} }} = 3\left( 1 \right)\\
\frac{{\left| {3a - 4b + c} \right|}}{{\sqrt {{a^2} + {b^2}} }} = 3\left( 2 \right)
\end{array} \right. \Rightarrow \frac{{\left| {2b + c} \right|}}{{\sqrt {{a^2} + {b^2}} }} = \frac{{\left| {3a - 4b + c} \right|}}{{\sqrt {{a^2} + {b^2}} }} \Leftrightarrow \left| {2b + c} \right| = \left| {3a - 4b + c} \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
3a - 4b + c = 2b + c\\
3a - 4b + c =  - 2b - c
\end{array} \right.
\end{array}\)

\( \Leftrightarrow \left[ \begin{array}{l}
a = 2b\\
3a - 2b + 2c = 0
\end{array} \right.\). Mặt khác từ (1): \({\left( {2b + c} \right)^2} = 9\left( {{a^2} + {b^2}} \right) \Leftrightarrow \)

- Trường hợp: thay a = 2b vào (1):

\({\left( {2b + c} \right)^2} = 9\left( {4{b^2} + {b^2}} \right) \Leftrightarrow 41{b^2} - 4bc - {c^2} = 0.\Delta {'_b} = 4{c^2} + 41{c^2} = 45{c^2} \Leftrightarrow \left[ \begin{array}{l}
b = \frac{{2b - 3\sqrt 5 c}}{4}\\
b = \frac{{\left( {2 + 3\sqrt 5 } \right)c}}{4}
\end{array} \right.\)

- Do đó ta có hai đường thẳng cần tìm :

\(\begin{array}{l}
{d_1}:\frac{{\left( {2 - 3\sqrt 5 } \right)}}{2}x + \frac{{\left( {2 - 3\sqrt 5 } \right)}}{4}y + 1 = 0 \Leftrightarrow 2\left( {2 - 3\sqrt 5 } \right)x + \left( {2 - 3\sqrt 5 } \right)y + 4 = 0\\
{d_1}:\frac{{\left( {2 + 3\sqrt 5 } \right)}}{2}x + \frac{{\left( {2 + 3\sqrt 5 } \right)}}{4}y + 1 = 0 \Leftrightarrow 2\left( {2 + 3\sqrt 5 } \right)x + \left( {2 + 3\sqrt 5 } \right)y + 4 = 0
\end{array}\)

- Trường hợp: \(c = \frac{{2b - 3a}}{2}\), thay vào (1): \(\frac{{\left| {2b + \frac{{2b - 3a}}{2}} \right|}}{{\sqrt {{a^2} + {b^2}} }} = 3 \Leftrightarrow \left| {2b - a} \right| = \sqrt {{a^2} + {b^2}} \)

\( \Leftrightarrow {\left( {2b - a} \right)^2} = {a^2} + {b^2} \Leftrightarrow 3{b^2} - 4ab = 0 \Leftrightarrow \left[ \begin{array}{l}
b = 0 \to c =  - \frac{a}{2}\\
b = \frac{{4a}}{3} \to c =  - \frac{a}{6}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
b = 0,a =  - 2c\\
b = \frac{{4a}}{3},a =  - 6c
\end{array} \right.\)

- Vậy có 2 đường thẳng \({d_3}:2x - 1 = 0,{d_4}:6x + 8y - 1 = 0\)