Đường tròn tâm \(I\left(-5;4\right)\) bán kính \(R=2\sqrt{10}\)
Ta có: \(S_{IAB}=\frac{1}{2}IA.IB.sin\widehat{AIB}=\frac{1}{2}R^2.sin\widehat{AIB}\le\frac{1}{2}R^2\)
\(\Rightarrow S_{max}\) khi \(sin\widehat{AIB}=1\Leftrightarrow AI\perp BI\Rightarrow AB=R\sqrt{2}=4\sqrt{5}\)
Khi đó \(MAIB\) là hình vuông
\(\Rightarrow IM=AB=4\sqrt{5}\)
Do M thuộc d nên tọa độ có dạng: \(M\left(m;m+5\right)\Rightarrow\overrightarrow{IM}=\left(m+5;m+1\right)\)
\(\Rightarrow\left(m+5\right)^2+\left(m+1\right)^2=80\)
\(\Leftrightarrow m^2+6m-27=0\Rightarrow\left[{}\begin{matrix}m=3\\m=-9\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}M\left(3;8\right)\\M\left(-9;-4\right)\end{matrix}\right.\)
b/ Gọi \(P\left(a;a+5\right)\Rightarrow\overrightarrow{IP}=\left(a+5;a+1\right)\)
Ta có: \(S_{PAI}=\frac{1}{2}AI.AP=\frac{1}{2}R.\sqrt{IP^2-R^2}=3\sqrt{10}\)
\(\Leftrightarrow\sqrt{10}.\sqrt{IP^2-40}=3\sqrt{10}\)
\(\Leftrightarrow IP^2=49\Leftrightarrow\left(a+5\right)^2+\left(a+1\right)^2=49\)
\(\Leftrightarrow2a^2+12a-23=0\Rightarrow a=\frac{-6\pm\sqrt{82}}{2}\Rightarrow P...\)
