Tâm I (1;-1)

vecto IA(-3;4)

=> IA = R =\(\sqrt{3^2+4^2}=5\)

=>pt: \(\left(x-1\right)^2+\left(y+1\right)^2=25\)

Gọi phương trình đường tròn \(\left(C\right):\left(x-a\right)^2+\left(y-b\right)^2=R^2\)

Gọi \(I\) là trung điểm \(AB\)

\(\Rightarrow I\left(1;-1\right)\), đồng thời \(I\) cũng là tâm đường tròn \(\left(C\right)\)

\(R=IA=\sqrt{\left(1+2\right)^2+\left(-1-3\right)^2}=5\)

\(\Rightarrow\left(C\right):\left(x-1\right)^2+\left(y+1\right)^2=25\)

Đề bài yeu cầu gì?

a) có \(\widehat{xOy}+\widehat{yOz}=180^o\left(kb\right)\)

\(hay70^o+\widehat{yOz}=180^o\)

\(\Rightarrow\widehat{yOz}=180^o-70^o\)

\(\Rightarrow\widehat{yOz}=110^o\)

b) có \(\widehat{xOy}< \widehat{xOt}\left(70^o< 120^o\right)\)

=> Oy nằm giữa hai tia Ox và Ot

\(\Rightarrow\widehat{xOy}+\widehat{yOt}=\widehat{xOt}\)

\(hay70^o+\widehat{yOt}=120^o\)

\(\Rightarrow\widehat{yOt}=120^o-70^o\)

\(\Rightarrow\widehat{yOt}=50^o\)

Cùng nhé !

\(\left\{{}\begin{matrix}\overrightarrow{MB}.\overrightarrow{MC}=0\\MB=MC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[\left(x_B-x\right)\overrightarrow{i}+\left(y_B-y\right)\overrightarrow{j}\right]\left[\left(x_c-x\right)\overrightarrow{i}+\left(y_C-y\right)\overrightarrow{j}\right]=0\\\sqrt{\left(x_B-x\right)^2+\left(y_B-y\right)^2}=\sqrt{\left(x_C-x\right)^2+\left(y_C-y\right)^2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(4-x\right)\left(-3-x\right)+\left(-2-y\right)\left(-1-y\right)=0\\\left(4-x\right)^2+\left(-2-y\right)^2=\left(-3-x\right)^2+\left(-1-y\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-x+3y-10=0\\y+5=7x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x-1\right)=0\\y=7x-5\end{matrix}\right.\)

\(\Rightarrow\)M(x;y): (0;-5) ; (1;2)

B^AC'

B^AC' là sao à bạn