Đường tròn (C) tâm \(I\left(1;-2\right)\) bán kính \(R=3\)
\(S_{IAB}=\dfrac{1}{2}IA.IB.sin\widehat{AIB}=\dfrac{1}{2}R^2.sin\widehat{AIB}\le\dfrac{1}{2}R^2\)
\(S_{max}\) khi \(sin\widehat{AIB}=1\Rightarrow\Delta AIB\) vuông cân tại I
\(\Rightarrow AB=R\sqrt{2}=3\sqrt{2}\)
\(\Rightarrow d\left(I;AB\right)=\dfrac{AB}{2}=\dfrac{3\sqrt{2}}{2}\)
Gọi phương trình AB có dạng: \(a\left(x+1\right)+b\left(y+3\right)=0\) với a;b ko đồng thời bằng 0
\(d\left(I;AB\right)=\dfrac{\left|a-2b+a+3b\right|}{\sqrt{a^2+b^2}}=\dfrac{3\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}\left|2a+b\right|=3\sqrt{a^2+b^2}\)
\(\Leftrightarrow2\left(4a^2+4ab+b^2\right)=9\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2-8ab+7b^2=0\Rightarrow\left[{}\begin{matrix}a=b\\a=7b\end{matrix}\right.\)
Chọn b=1 \(\Rightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(1;1\right)\\\left(a;b\right)=\left(1;7\right)\end{matrix}\right.\)
Có 2 đường thẳng thỏa mãn: \(\left[{}\begin{matrix}1\left(x+1\right)+1\left(y+3\right)=0\\1\left(x+1\right)+7\left(y+3\right)=0\end{matrix}\right.\)
