\(x^2+y^2-2x+6y+1=0\)
Ta có :
\(\left\{{}\begin{matrix}-2a=-2\\-2b=6\\c=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-3\\c=1\end{matrix}\right.\)
\(\Rightarrow I\left(a;b\right)\Rightarrow I\left(1;-3\right)\)
Bán kính \(R=\sqrt{a^2+b^2-c}=\sqrt{1^2+\left(-3\right)^2-1}=3\)
Đúng 1
Bình luận (0)
