Đáp án:
$\begin{array}{l}
\left( C \right):{x^2} + {y^2} - 2x + 6y + 5 = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} + {\left( {y + 3} \right)^2} = 5\\
\Leftrightarrow I\left( {1; - 3} \right);R = \sqrt 5 \\
\Delta //d\\
\Leftrightarrow \Delta :y = - 2x + b\left( {b\# 1} \right)\\
hay\,\Delta :2x + y - b = 0\\
\Leftrightarrow {d_{I - \Delta }} = R\\
\Leftrightarrow \dfrac{{\left| {2.1 + \left( { - 3} \right) - b} \right|}}{{\sqrt {{2^2} + {1^2}} }} = \sqrt 5 \\
\Leftrightarrow \left| {b + 1} \right| = 5\\
\Leftrightarrow \left[ \begin{array}{l}
b = 4\left( {tm} \right)\\
b = - 6\left( {tm} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\Delta :y = - 2x + 4\\
\Delta :y = - 2x - 6
\end{array} \right.\\
+ Khi:y = - 2x + 4\\
Thay\,vao\,\left( C \right):\\
{\left( {x - 1} \right)^2} + {\left( {y + 3} \right)^2} = 5\\
\Leftrightarrow {\left( {x - 1} \right)^2} + {\left( { - 2x + 4 + 3} \right)^2} = 5\\
\Leftrightarrow 5{x^2} - 30x + 50 = 5\\
\Leftrightarrow {x^2} - 6x + 9 = 0\\
\Leftrightarrow x = 3\\
\Leftrightarrow y = - 2x + 4 = - 2
\end{array}$
=> tiếp điểm là B(3;-2)
$\begin{array}{l}
+ Khi:\Delta :y = - 2x - 6\\
Thay\,vao\,\left( C \right):\\
{\left( {x - 1} \right)^2} + {\left( {y + 3} \right)^2} = 5\\
\Leftrightarrow {\left( {x - 1} \right)^2} + {\left( { - 2x - 6 + 3} \right)^2} = 5\\
\Leftrightarrow 4{x^2} + 10x + 10 = 5\\
\Leftrightarrow 4{x^2} + 10x + 5 = 0\\
\Leftrightarrow x = \dfrac{{ - 5 \pm \sqrt 5 }}{4} \Leftrightarrow y = \dfrac{{ - 7 \pm \sqrt 5 }}{2}\\
\Leftrightarrow TD:\left( {\dfrac{{ - 5 \pm \sqrt 5 }}{4};\dfrac{{ - 7 \pm \sqrt 5 }}{2}} \right)
\end{array}$
