$S_1=x^2+y^2+2 x+4 y-4=0, \quad O_1(-1,-2), \quad \ r_1=3 $
$S_2=x^2+y^2+6 y=0,\quad O_2(0,-3), \quad r_2=3 $
$\left|r_1-r_2\right| \quad < \quad d_{O_1,O_2}=\sqrt{2} \quad < \quad6 $
$S=S_1+k S_2=0 \text { where } k \text { is real and } k\neq{-1}$
Solution:
$S=\left(x^2+y^2+2 x+4 y-4\right)+k\left(x^2+y^2+6 y\right)=0 $
$(1+k) x^2+(1+k) y^2+2 x+(4+6k)y -4 =0$