\(x^2=2mx+1\Leftrightarrow x^2-2mx-1=0\Rightarrow\Delta'>0\Leftrightarrow m^2+1>0\left(luônđúng\right)\)
\(\Rightarrow\left(P\right)\left(d\right)\) \(luôn\) \(cắt\) \(tại2\) \(điểm\) \(pbA;B\Rightarrow\left\{{}\begin{matrix}x_A+x_B=2m\\xa.xb=-1\end{matrix}\right.\)
\(I\) \(trunng\) \(điểmAB\Rightarrow I\left(\dfrac{x_A+x_B}{2};\dfrac{y_A+y_B}{2}\right)=\left(\dfrac{2m}{2};\dfrac{2mx_A+1+2mx_B+1}{2}\right)=\left(m;m.x_A+mx_B+1\right)\)
\(\Rightarrow OI=\sqrt{10}=\sqrt{m^2+\left(mx_A+mx_B+1\right)^2}\)
\(\Leftrightarrow10=m^2+\left[m\left(x_A+x_B\right)+1\right]^2=m^2+\left(2m^2+1\right)^2\)
\(\Leftrightarrow m^2+4m^4+4m^2+1=10\Leftrightarrow4m^4+5m^2-9=0\)
\(đặt:m^2=t\ge0\Rightarrow4t^2+5t-9=0\Leftrightarrow\left[{}\begin{matrix}t=1\left(tm\right)\Rightarrow m=\pm1\\t=-\dfrac{9}{4}\left(ktm\right)\end{matrix}\right.\)
