$(C)$: tâm $I$, bán kính $R=2$.
$Q_{(O;45^o)}: (C)\to (C')$
$V_{(O;\sqrt2)}: (C')\to (C'')$
Qua phép đồng dạng, bán kính ảnh là:
$R''=|k|R=2\sqrt2$
Ta có $\sin45^o=\cos45^o=\dfrac{1}{\sqrt2}$
$\Rightarrow I'(1.\dfrac{1}{\sqrt2}-1.\dfrac{1}{\sqrt2}; 1.\dfrac{1}{\sqrt2}+1.\dfrac{1}{\sqrt2})=(0;\sqrt2)$
$\vec{OI'}(0;\sqrt2)$
$\Rightarrow \vec{OI''}=\sqrt2\vec{OI'}=(0;2)$
$\Rightarrow I''(0;2)$
Vậy $(C''): x^2+(y-2)^2=(2\sqrt2)^2=8$
