Do \(A\left(a;b\right)\in d\Rightarrow3a-2b-1=0\)
\(\Leftrightarrow3a-2b=1\)
\(\Rightarrow1=\left(3a-2b\right)^2\le\left(9+4\right)\left(a^2+b^2\right)\)
\(\Rightarrow a^2+b^2\ge\frac{1}{13}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}3a-2b=1\\\frac{a}{3}=\frac{b}{-2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{3}{13}\\b=-\frac{2}{13}\end{matrix}\right.\)