a,
$D_B: A\to A'$
$\to A'(-2.4-5; 2.3+2)=(-13; 8)$
$D_A: B\to B'$
$\to B'(2.5+4; -2.2-3)=(14; -7)$
b,
$D_A: d\to d'$
Chọn điểm $P(0;0,5), Q(1,5; 1)\in d$
$\to \begin{cases} P'(2.5-0; -2.2-0,5)=(10; -4,5)\\ Q'(2.5-1,5; -2.2-1)=(8,5; -5)\end{cases}$
$\vec{u_{d'}}=\vec{P'Q'}(-1,5; -0,5)$
$\to \vec{n_{d'}}(1;-3)$
$d': x-8,5-3(y+5)=0$
$\to 2x-6y-47=0$
c,
$D_A: (C)\to (C')$
$(C)$: tâm $K(-2;-1); R^2=4$
$\to R'^2=4$
$K'(2.5+2; -2.2+1)=(12; -3)$
Vậy $(C'): (x-12)^2+(y+3)^2=4$