What you need to do is write the equation of the circle in standard form,
[tex](x-a)^2+(y-b)^2=r^2[/tex]
so that it's easy to identify the two central traits of the circle, its center [tex](a,b)[/tex] and its radius [tex]r[/tex].
By completing the square, we have
[tex]x^2+3x+y^2+2y-\dfrac34=0[/tex]
[tex]\left(x^2+3x+\dfrac94\right)+\left(y^2+2y+1\right)=\dfrac34+\dfrac94+1[/tex]
[tex]\left(x+\dfrac32\right)^2+(y+1)^2=4=2^2[/tex]
and from this we can tell the circle is centered at [tex]\left(-\dfrac32,-1\right)[/tex] with radius 2.
The line [tex]y=k[/tex] is horizontal and will intersect the circle whenever [tex]-1-2\le k\le-1+2[/tex], or [tex]-3\le k\le1[/tex]. So the line will not intersect the circle if [tex]k<-3[/tex] or [tex]k>1[/tex].
