$(C)$: tâm $I(2;-4)$, $R=\sqrt{2^2+4^2-18}=\sqrt2$
Gọi tiếp tuyến là $d: a(x-1)+b(y-1)=0\to ax+by-a-b=0$
Ta có: $d(I;d)=R$
$\to \dfrac{|2a-4b-a-b|}{\sqrt{a^2+b^2}}=\sqrt2$
$\to |a-5b|=\sqrt2.\sqrt{a^2+b^2}$
$\to a^2-10ab+25b^2=2a^2+2b^2$
$\to a^2+10ab-23b^2=0$
$\to \Big(\dfrac{a}{b}\Big)^2+10.\dfrac{a}{b}-23=0$
$\to \dfrac{a}{b}=-5\pm4\sqrt3$
Chọn $b=1\to a=-5\pm4\sqrt3$
Vậy $d: (-5\pm4\sqrt3)(x-1)+y-1=0$
