\(AB=\sqrt{\left(-2-2\right)^2+\left(4-1\right)^2}=5\)
\(OA=\sqrt{2^2+1^2}=\sqrt{5}\)
\(OB=\sqrt{\left(-2\right)^2+4^2}=2\sqrt{5}\)
Ta có \(OA^2+OB^2=25=AB^2\Rightarrow\Delta OAB\) vuông tại O
\(\Rightarrow\frac{1}{OH^2}=\frac{1}{OA^2}+\frac{1}{OB^2}=\frac{1}{5}+\frac{1}{20}=\frac{1}{4}\Rightarrow OH=2\)
\(\Rightarrow\sqrt{x_H^2+y_H^2}=2\Leftrightarrow x_H^2+y_H^2=4\)
Lại có \(\left\{{}\begin{matrix}HA=\frac{OA^2}{AB}=1\\HB=\frac{OB^2}{AB}=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\left(2-x_H\right)^2+\left(1-y_H\right)^2}=1\\\sqrt{\left(-2-x_H\right)^2+\left(4-y_H\right)^2}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_H^2+y_H^2-4x_H-2y_H+5=1\\x_H^2+y^2_H+4x_H-8y_H+20=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x_H+y_H=4\\x_H-2y_H=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x_H=\frac{6}{5}\\y_H=\frac{8}{5}\end{matrix}\right.\)
\(\Rightarrow H\left(\frac{6}{5};\frac{8}{5}\right)\)
