$V_{(I;k)}:(C) \to (C')$
a,
(đề thiếu bán kính $(C)$ )
b,
$(C)$: tâm $M(0;0)$, $R=\sqrt1=1$
$\to R'=|k|R=4$
$\to R'^2=4^2=16$
$\vec{IM}(-3; 1)$
$\to \vec{IM'}=-4\vec{IM}=(-3.(-4); 1.(-4))=(12; -4)$
$\to M'(12+3; -4-1)=(15;-5)$
Vậy $(C')$: $(x-15)^2+(y+5)^2=16$