2.
下面给出四个命题的表述:
\(①\)直线\((3+m)x+4y-3+3m=0(m∈R)\)恒过定点\((-3,3)\);
\(②\)线段\(AB\)的端点\(B\)的坐标是\((3,4)\),\(A\)在圆\(x^{2}+y^{2}=4\)上运动,则线段\(AB\)的中点\(M\)的轨迹方程\({{\left( x-\dfrac{3}{2} \right)}^{2}}+{{(y-2)}^{2}}=1\);
\(③\)已知\(M=\left\{ \left.\left(x,y\right) \right|y= \sqrt{1-{x}^{2}}\right\} \),\(N=\{(x,y)|y=x+b\}\),若\(M∩N\neq \varnothing \),则\(b∈\left[- \sqrt{2}, \sqrt{2}\right] \);
\(④\)已知圆\(C:(x-b)^{2}+(y-c)^{2}=a^{2}(a > 0,b > 0,c > 0)\)与\(x\)轴相交,与\(y\)轴相离,则直线\(ax+by+c=0\)与直线\(x+y+1=0\)的交点在第二象限.
其中表述正确的是 \((\) \()\)