Đáp án:
a) Gọi tâm đường tròn là: I (x;y)
VÌ I nằm trên đường thẳng nên: 3x-y-11=0 => y=3x-11
=> I (x; 3x-11)
A,B nằm trên đường tròn nên:
$\begin{array}{l}
IA = IB\\
\Rightarrow I{A^2} = I{B^2}\\
\Rightarrow {\left( {x - 3} \right)^2} + {\left( {3x - 11 - 2} \right)^2} = {\left( {x - 1} \right)^2} + {\left( {3x - 11 + 1} \right)^2}\\
\Rightarrow {x^2} - 6x + 9 + 9{x^2} - 78x + 169\\
= {x^2} - 2x + 1 + 9{x^2} - 60x + 100\\
\Rightarrow 22x = 77\\
\Rightarrow x = \frac{7}{2}\\
\Rightarrow I\left( {\frac{7}{2}; - \frac{1}{2}} \right)\\
\Rightarrow R = IA = \frac{{\sqrt {26} }}{2}\\
\Rightarrow \left( C \right):{\left( {x - \frac{7}{2}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} = \frac{{13}}{2}\\
b)d//d':x - y - 11 = 0\\
\Rightarrow d:x - y + c = 0\left( {c \ne - 11} \right)\\
Tâm\,I\left( { - 1;4} \right);R = \sqrt {18} \\
d\,là\,TT\left( {C'} \right)\\
\Rightarrow {d_{I - d}} = R\\
\Rightarrow \frac{{\left| { - 1 - 4 + c} \right|}}{{\sqrt {1 + 1} }} = \sqrt {18} \\
\Rightarrow {\left( {c - 5} \right)^2} = 36\\
\Rightarrow \left[ \begin{array}{l}
c - 5 = 6\\
c - 5 = - 6
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
c = 11\left( {tm} \right)\\
c = - 1\left( {tm} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
d:x - y + 11 = 0\\
d:x - y - 1 = 0
\end{array} \right.
\end{array}$
