Đáp án:
$(C):\,x^2+(y-1)^2=2$
Giải thích các bước giải:
$Δ:\,x-y-1=0$
Gọi $I(a;b)$ là tâm đường tròn $(C)$
$⇒AI=BI=d_{(I,Δ)}=R$
$\overrightarrow{AI}=(a+1;b)⇒AI=\sqrt{(a+1)^2+b^2}$
$\overrightarrow{BI}=(a-1;b-2)⇒BI=\sqrt{(a-1)^2+(b-2)^2}$
$AI=BI$
$⇒\sqrt{(a+1)^2+b^2}=\sqrt{(a-1)^2+(b-2)^2}$
$⇒(a+1)^2+b^2=(a-1)^2+(b-2)^2$
$⇒a^2+2a+1+b^2=a^2-2a+1+b^2-4b+4$
$⇒4a+4b=4$
$⇒b=1-a$
$AI=d_{(I,Δ)}$
$⇒\sqrt{(a+1)^2+b^2}=\dfrac{|a-b-1|}{\sqrt{1^2+1^2}}$
$⇒\sqrt{2}.\sqrt{(a+1)^2+(1-a)^2}=|a+a-1-1|$
$⇒2|a-1|=\sqrt{2}.\sqrt{2a^2+2}$
$⇒2|a-1|=2\sqrt{a^2+1}$
$⇒|a-1|=\sqrt{a^2+1}$
$⇒a^2-2a+1=a^2+1$
$⇒2a=0$
$⇒a=0⇒b=1$
$⇒I(0;1)$
$⇒R=\sqrt{(a+1)^2+b^2}=\sqrt{1^2+1^2}=\sqrt{2}$
$⇒(C):\,x^2+(y-1)^2=2$.
