Question

Cho (C): (x-3)²+(y-4)²=4. Cho B(4,1);C(8,3). Tìm A thuộc (C) sao cho tam giác ABC vuông tại A

Answer 1

Gọi \(A\left(x;y\right)\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{BA}=\left(x-4;y-1\right)\\\overrightarrow{CA}=\left(x-8;y-3\right)\end{matrix}\right.\)

\(\overrightarrow{BA}.\overrightarrow{CA}=0\Leftrightarrow\left(x-4\right)\left(x-8\right)+\left(y-1\right)\left(y-3\right)=0\)

\(\Leftrightarrow x^2-12x+y^2-4y+35=0\) (1)

Do A thuộc (C) nên ta cũng có:

\(\left(x-3\right)^2+\left(y-4\right)^2=4\Leftrightarrow x^2-6x+y^2-8y+21=0\) (2)

Trừ (1) cho (2): \(-6x+4y+14=0\Rightarrow x=\frac{2}{3}y+\frac{7}{3}\)

Thay vào (1):

\(\left(\frac{2y+7}{3}\right)^2-12\left(\frac{2y+7}{3}\right)+y^2-4y+35=0\)

\(\Leftrightarrow13y^2-80y+112=0\Rightarrow\left[{}\begin{matrix}y=4\Rightarrow x=5\\y=\frac{28}{13}\Rightarrow x=\frac{49}{13}\end{matrix}\right.\)