a, bán kính đường tròn là \(IA=\sqrt{\left(3-1\right)^2+\left(1-3\right)^2}=2\sqrt{2}\)
\(\Rightarrow\) pt đường tròn \(\left(C\right)\left(x-1\right)^2+\left(y-3\right)^2=\left(2\sqrt{2}\right)^2=8\)
b, bán kính đường tròn là :
\(R=\left(d;\Delta\right)=\dfrac{\left|4.\left(-1\right)-3.2+2\right|}{\sqrt{4^2}+\left(-3\right)^2}=\dfrac{8}{5}\)
\(\Rightarrow\) pt đường tròn \(\left(C\right)\left(x+1\right)^2+\left(y-2\right)^2=\dfrac{64}{25}\)
c, Gọi I là trung điểm AB => I là tâm đường tròn ( C)
\(\Rightarrow\left\{{}\begin{matrix}x_I=\dfrac{1+7}{2}=4\\y_I=\dfrac{1+5}{2}=3\end{matrix}\right.\) \(\Rightarrow I\left(4;3\right)\)
\(\overrightarrow{AB}=\left(6;4\right)\Rightarrow AB=2\sqrt{13}\Rightarrow R=\dfrac{AB}{2}=\dfrac{2\sqrt{13}}{2}=\sqrt{13}\)
pt đường tròn \(\left(C\right)\left(x-4\right)^2+\left(y-3\right)^2=13\)
d, gọi I là tâm ( C) và vectơ I có tọa độ ( a; b)
=> \(AI^2=BI^2\)
\(\Leftrightarrow\left(a+1\right)^2+\left(b-2\right)^2=\left(a+2\right)^2+\left(b-3\right)^2\)
\(\Leftrightarrow a^2+2a+1+b^2-4b+4=a^2+4a+4+b^2-6b+9\)
\(\Leftrightarrow2a-4b+5=4a-6b+13\)
\(\Leftrightarrow2a-2b+8=0\Rightarrow a-b=-8\left(1\right)\)
Vì I thuộc d => 3a-b=-10 ( 2)
Từ (1) và (2) suy ra : \(\left\{{}\begin{matrix}a=-1\\b=7\end{matrix}\right.\)
\(R=IA=\sqrt{\left(-1-1\right)^2+\left(2-7\right)^2}=\sqrt{29}\)
\(\Rightarrow pt\left(C\right)\left(x+1\right)^2+\left(y-7\right)^2=29\)
e, \(pt\left(C\right)\) có dạng \(x^2+y^2-2ax-2by+C=0\left(a^2+b^2-c>0\right)\)
\(\left\{{}\begin{matrix}A\left(1;2\right)\in\left(C\right)\\B\left(5;2\right)\in\left(C\right)\\C\left(1;-3\right)\in\left(C\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}1+4-2a-4y+C=0\\25+4-10a-4y+C=0\\1+9-2a+6y+C=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-\dfrac{1}{2}\\c=-1\end{matrix}\right.\)
\(\Rightarrow\left(C\right):x^2+y^2-6x+y-1=0\)
