Given the equation of the circle ;
[tex]x^2+y^2+10x=-30[/tex]Completing the square of the terms of x
so,
[tex]\begin{gathered} (x^2+10x+(\frac{10}{2})^2)+y^2=-30+(\frac{10}{2})^2 \\ \\ (x^2+10x+25)+y^2=-30+25 \\ \\ (x+5)^2+y^2=-5 \end{gathered}[/tex]The general equation of the circle is :
[tex](x-h)^2+(y-k)^2=r^2[/tex]Comparing the equations together :
[tex]\begin{gathered} r^2=-5 \\ \\ r=\sqrt[]{-5} \end{gathered}[/tex]so, the given equation represents a circle with a non-real radius.