Đường tròn tâm \(I\left(3;-1\right)\) bán kính \(R=\sqrt{3^2+\left(-1\right)^2-6}=2\)
a/ \(\overrightarrow{AI}=\left(2;-3\right)\Rightarrow AI=\sqrt{2^2+\left(-3\right)^2}=\sqrt{13}>2\)
\(\Rightarrow\) A nằm ngoài đường tròn
b/ Gọi tiếp tuyến d qua A có dạng:
\(a\left(x-1\right)+b\left(y-2\right)=0\Leftrightarrow ax+by-a-2b=0\) (\(a^2+b^2\ne0\))
d tiếp xúc (C) \(\Leftrightarrow d\left(I;d\right)=R\)
\(\Leftrightarrow\frac{\left|3a-b-a-2b\right|}{\sqrt{a^2+b^2}}=2\Leftrightarrow\left|2a-3b\right|=\sqrt{4a^2+4b^2}\)
\(\Leftrightarrow4a^2+9b^2-12ab=4a^2+4b^2\)
\(\Leftrightarrow5b^2-12ab=0\Rightarrow\left[{}\begin{matrix}b=0\\5b=12a\end{matrix}\right.\) \(\Rightarrow\left(a;b\right)=\left[{}\begin{matrix}\left(1;0\right)\\\left(5;12\right)\end{matrix}\right.\)
Có 2 tiếp tuyến thỏa mãn: \(\left[{}\begin{matrix}x-1=0\\5\left(x-1\right)+12\left(y-2\right)=0\end{matrix}\right.\)
c/ \(AT_1=\sqrt{IA^2-R^2}=3\) \(\Rightarrow S_{AT_1IT_2}=AT_1.R=6\)
Gọi H là trung điểm \(T_1T_2\Rightarrow\frac{1}{HT_1^2}=\frac{1}{AT_1^2}+\frac{1}{R^2}\Rightarrow HT_1=\frac{6\sqrt{13}}{13}\)
\(\Rightarrow IH=\sqrt{R^2-HT_1^2}=\frac{4\sqrt{13}}{13}\)
\(\Rightarrow S_{IT_1T_2}=2S_{IHT_1}=IH.HT_1=\frac{24}{13}\)
\(\Rightarrow S_{AT_1T_2}=6-\frac{24}{13}=\frac{54}{13}\)
