Đáp án:

\(d\left( {O;d} \right) = \frac{{3\sqrt 5 }}{5}\)

Giải thích các bước giải: Gọi phương trình đường thẳng đi qua M là y=k(x+3)+1=kx+3k+1 => kx-y+3k+1=0 (\(\Delta\)) Đường tròn (C) có tâm I(1;3) bán kính R=2. (d) tiếp xúc với (C) => d(I;d)=R=2 \(\begin{array}{l} \frac{{\left| {k - 3 + 3k + 1} \right|}}{{\sqrt {{k^2} + 1} }} = 2 \Leftrightarrow \left| {4k - 2} \right| = 2\sqrt {{k^2} + 1} \\ \Leftrightarrow 16{k^2} - 16k + 4 = 4{k^2} + 4\\ \Leftrightarrow 12{k^2} - 16k = 0 \Leftrightarrow k = 0\,\,hoac\,\,k = \frac{4}{3}\\ \Rightarrow \Delta :\,\,y = 1\,\,hoac\,\,\Delta :\,\,y = \frac{4}{3}x + 5\\ y = 1 \Rightarrow {x^2} - 2x + 1 = 0 \Leftrightarrow x = 1 \Rightarrow A\left( {1;1} \right)\\ Pt\,\,AB\,\,di\,\,qua\,\,\,A\,\,va\,\,nhan\,\,\overrightarrow {IM} = \left( { - 4; - 2} \right) = - 2\left( {2;1} \right)\,\,la\,\,1\,\,VTPT\\ \Rightarrow AB:\,\,2\left( {x - 1} \right) + 1\left( {y - 1} \right) = 0 \Leftrightarrow 2x + y - 3 = 0\\ Lay\,\,M\left( {x;y} \right) \in AB,\,\,goi\,\,M'\left( {x';y'} \right) = {Q_{\left( {O;{{45}^0}} \right)}}\left( M \right)\\ \Rightarrow \left\{ \begin{array}{l} x' = x\frac{{\sqrt 2 }}{2} - y\frac{{\sqrt 2 }}{2}\\ y' = x\frac{{\sqrt 2 }}{2} + y\frac{{\sqrt 2 }}{2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = \frac{{\sqrt 2 }}{2}x' + \frac{{\sqrt 2 }}{2}y'\\ y = - \frac{{\sqrt 2 }}{2}x' + \frac{{\sqrt 2 }}{2}y' \end{array} \right.\\ \Rightarrow M\left( {\frac{{\sqrt 2 }}{2}x' + \frac{{\sqrt 2 }}{2}y'; - \frac{{\sqrt 2 }}{2}x' + \frac{{\sqrt 2 }}{2}y'} \right) \in AB\\ \Rightarrow 2\left( {\frac{{\sqrt 2 }}{2}x' + \frac{{\sqrt 2 }}{2}y'} \right) + \left( { - \frac{{\sqrt 2 }}{2}x' + \frac{{\sqrt 2 }}{2}y'} \right) - 3 = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}x' + \sqrt 2 y' - 3 = 0\\ \Leftrightarrow \sqrt 2 x' + 2\sqrt 2 y' - 3\sqrt 2 = 0\\ \Rightarrow d:\,\,\sqrt 2 x' + 2\sqrt 2 y' - 3\sqrt 2 = 0\\ \Rightarrow d\left( {O;d} \right) = \frac{{\left| { - 3\sqrt 2 } \right|}}{{\sqrt {2 + 8} }} = \frac{{3\sqrt 2 }}{{\sqrt {10} }} = \frac{{3\sqrt 5 }}{5} \end{array}\)