$T_{\overrightarrow{u}(1;1)}: (C)\to (C')$
$(C)$: tâm $I(2;6)$
$\to I'(2+1; 6+1)=(3;7)$
$D_{d: x+y-1=0}: (C')\to (C'')$
$ R''=R'=R=\sqrt9=3$
$\to R''^2=3^2=9$
Gọi $\Delta: \begin{cases} I'\in \Delta\\ \Delta\bot d\end{cases}$
$\to \Delta: x-y+c=0$
$I'\in \Delta\to 3-7+c=0$
$\to c=4$
$\to \Delta: x-y+4=0$
Gọi $M=\Delta\cap d\to M(2,5; -1,5)$
$I'I''$ có trung điểm là $M$
$\to I''(2.2,5-2; -2.1,5-7)=(3; -10)$
Vậy $(C''): (x-3)^2+(y+10)^2=9$