Question

[HUỲNH NGỌC BẢO ÂN]

Viết PT đường tròn (c) trog trường hợp sau:

đi qua A(1;1) B(1;4) và tiếp xúc trục Ox

Answer 1

[Nguyễn Việt Lâm]

Gọi tâm đường tròn là \(I\left(a;b\right)\Rightarrow IA=IB=d\left(I;Ox\right)=b\)

\(\left\{{}\begin{matrix}\overrightarrow{AI}=\left(a-1;b-1\right)\\\overrightarrow{BI}=\left(a-1;b-4\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}AI^2=\left(a-1\right)^2+\left(b-1\right)^2\\BI^2=\left(a-1\right)^2+\left(b-4\right)^2\end{matrix}\right.\)

\(AI^2=BI^2\Rightarrow\left(b-1\right)^2=\left(b-4\right)^2\)

\(\Rightarrow-2b+1=-8b+16\Rightarrow b=\dfrac{5}{2}\)

Lại có:

\(IA=b\Rightarrow IA^2=b^2\Rightarrow\left(a-1\right)^2+\left(\dfrac{5}{2}-1\right)^2=\left(\dfrac{5}{2}\right)^2\)

\(\Rightarrow\left(a-1\right)^2=4\Rightarrow\left[{}\begin{matrix}a=3\\a=-1\end{matrix}\right.\)

Có 2 đường tròn thỏa mãn: \(\left[{}\begin{matrix}\left(x-3\right)^2+\left(y-\dfrac{5}{2}\right)^2=\dfrac{25}{4}\\\left(x+1\right)^2+\left(y-\dfrac{5}{2}\right)^2=\dfrac{25}{4}\end{matrix}\right.\)