Đáp án:
$\begin{array}{l}
a){x^2} + {y^2} + 4x - 2y - 1 = 0\\
\Rightarrow {\left( {x + 2} \right)^2} + {\left( {y - 1} \right)^2} = 6\\
\Rightarrow Tam\,I\left( { - 2;1} \right);R = \sqrt 6 \\
b)PTTT//\Delta :3x - 4y + 2020 = 0\\
\Rightarrow 3x - 4y + d = 0\left( {d \ne 2020} \right)\\
\Rightarrow {d_{I - PTTT}} = R\\
\Rightarrow \frac{{\left| {3.\left( { - 2} \right) - 4.1 + d} \right|}}{{\sqrt {{3^2} + {{\left( { - 4} \right)}^2}} }} = \sqrt 6 \\
\Rightarrow \frac{{\left| {d - 10} \right|}}{5} = \sqrt 6 \\
\Rightarrow \left| {d - 10} \right| = 5\sqrt 6 \\
\Rightarrow \left[ \begin{array}{l}
d = 10 + 5\sqrt 6 \left( {tmdk} \right)\\
d = 10 - 5\sqrt 6 \left( {tmdk} \right)
\end{array} \right.\\
\Rightarrow PTTT:\left[ \begin{array}{l}
3x - 4y + 10 + 5\sqrt 6 = 0\\
3x - 4y + 10 - 5\sqrt 6 = 0
\end{array} \right.
\end{array}$
