Đáp án:
$\begin{array}{l}
1b)\left( d \right):y = - 3x + 3 \Rightarrow 3x + y - 3 = 0\\
R = {d_{I - d}} = \frac{{\left| {3.2 + 7 - 3} \right|}}{{\sqrt {{3^2} + 1} }} = \frac{{10}}{{\sqrt {10} }} = \sqrt {10} \\
\Rightarrow \left( C \right):{\left( {x - 1} \right)^2} + {\left( {y - 7} \right)^2} = {R^2} = 10\\
2):\\
\left( C \right):{x^2} + {y^2} + 4x + 4y - 1 = 0\\
\Rightarrow {\left( {x + 2} \right)^2} + {\left( {y + 2} \right)^2} = 9 = {3^2}\\
\Rightarrow I\left( { - 2; - 2} \right);R = 3\\
\Delta ':3x - 4y + d = 0\left( {d \ne - 2} \right)\\
{d_{I - \Delta '}} = \frac{{\left| {3.\left( { - 2} \right) - 4.\left( { - 2} \right) + d} \right|}}{{\sqrt {{3^2} + {4^2}} }} = \frac{{\left| {d + 2} \right|}}{5}\\
Do:{d^2} + {\left( {\frac{{AB}}{2}} \right)^2} = {R^2}\\
\Rightarrow \frac{{{{\left( {d + 2} \right)}^2}}}{{25}} + 5 = 9\\
\Rightarrow {\left( {d + 2} \right)^2} = 100\\
\Rightarrow \left[ \begin{array}{l}
d + 2 = 10\\
d + 2 = - 10
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
d = 8\\
d = - 12
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\Delta ':3x - 4y + 8 = 0\\
\Delta ':3x - 4y - 12 = 0
\end{array} \right.
\end{array}$
