If one of the circles $x^2+y^2+2ax+c=0$ and 

$x^2+y^2+2bx+c=0$ lies within the other, then

 

The centres of the two circles are $(- a, 0)$ and $(-b, 0)$ respectively and their radical axis is

$2x(a-b)=0\text{ i.c. }x=0$

Since one of the circles lies within the other. So, their points of intersection are imaginary, and hence the radical axis does not intersect them. Thus, both the circles lie on the same side 9f radical axis i.e. $x = 0$. Hence, a and b are of the same sign i.e. $ab> 0$. Also, since the origin lies on the radical axis $x = 0$. So, (0, 0) is outside the two circles. Therefore, $c > 0.$