Đường tròn (C) tâm \(I\left(0;0\right)\) bán kính R=1
Đường tròn \(\left(C_m\right)\) tâm \(I'\left(m+1;-2m\right)\) bán kính \(R'=\sqrt{5m^2+2m+6}\)
Ta có: \(II'=\sqrt{\left(m+1\right)^2+\left(2m\right)^2}=\sqrt{5m^2+2m+1}\)
Hai đường tròn tiếp xúc nhau khi:
\(\left[{}\begin{matrix}II'=R+R'\\II'=\left|R-R'\right|\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{5m^2+2m+1}=\sqrt{5m^2+2m+6}+1\left(vn\right)\\\sqrt{5m^2+2m+1}=\sqrt{5m^2+2m+6}-1\end{matrix}\right.\)
\(\Rightarrow\sqrt{5m^2+2m+1}+1=\sqrt{5m^2+2m+6}\)
\(\Leftrightarrow\sqrt{5m^2+2m+1}=2\)
\(\Leftrightarrow5m^2+2m-3=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=\dfrac{3}{5}\end{matrix}\right.\)
