CHÚC BẠN HỌC TỐT!!!
Trả lời:
Gọi $I(a;b)$ là tâm đường tròn $(C)$
$I\in Δ⇒I$ có dạng $I(2b-1;b)$
Ta có:
$\overrightarrow{AI}=(2b-2;b+2)⇒AI=\sqrt{(2b-2)^2+(b+2)^2}$
$\overrightarrow{BI}=(2b-4;b-1)⇒BI=\sqrt{(2b-4)^2+(b-1)^2}$
$AI=BI=R$
$⇒\sqrt{(2b-2)^2+(b+2)^2}=\sqrt{(2b-4)^2+(b-1)^2}$
$⇒(2b-2)^2+(b+2)^2=(2b-4)^2+(b-1)^2$
$⇒4b^2-8b+4+b^2+4b+4=4b^2-16b+16+b^2-2b+1$
$⇒5b^2-4b+8=5b^2-18b+17$
$⇒14b=9$
$⇒b=\dfrac{9}{14}$
$⇒I\bigg{(}\dfrac{2}{7};\dfrac{9}{14}\bigg{)}$
$⇔R=AI=\sqrt{\bigg{(}2.\dfrac{9}{14}-4\bigg{)}^2+\bigg{(}\dfrac{9}{14}-1\bigg{)}^2}=\sqrt{\dfrac{1469}{196}}$
$⇒(C):\,\bigg{(}x-\dfrac{2}{7}\bigg{)}^2+\bigg{(}y-\dfrac{9}{14}\bigg{)}^2=\dfrac{1469}{196}$.
