\(x=2\) là TCĐ, \(y=1\) là TCN \(\Rightarrow I\left(2;1\right)\)
\(y'=\dfrac{-4}{\left(x-2\right)^2}\)
Gọi hoành độ tiếp điểm là \(a\Rightarrow y=-\dfrac{4}{\left(a-2\right)^2}\left(x-a\right)+\dfrac{a+2}{a-2}\) là tiếp tuyến
\(x_A=2\Rightarrow y_A=-\dfrac{4}{\left(a-2\right)^2}\left(2-a\right)+\dfrac{a+2}{a-2}=\dfrac{a+6}{a-2}\) \(\Rightarrow A\left(2;\dfrac{a+6}{a-2}\right)\)
\(y_B=1\Rightarrow-\dfrac{4}{\left(a-2\right)^2}\left(x_A-a\right)+\dfrac{a+2}{a-2}=1\Rightarrow x_A=2a-2\) \(\Rightarrow B\left(2a-2;1\right)\)
\(\Rightarrow\overrightarrow{AB}=\left(2a-4;-\dfrac{8}{a-2}\right)\Rightarrow AB=\sqrt{4\left(a-2\right)^2+\dfrac{64}{\left(a-2\right)^2}}\)
\(AB=2\sqrt{\left(a-2\right)^2+\dfrac{16}{\left(a-2\right)^2}}\ge2\sqrt{2\sqrt{\dfrac{16\left(a-2\right)^2}{\left(a-2\right)^2}}}=4\sqrt{2}\)
\(\Rightarrow R=\dfrac{AB}{2}\ge2\sqrt{2}\)
\(\Rightarrow C=2\pi R\ge4\pi\sqrt{2}\)
