The circle $x^2 +y ^2 +2b_1x+c=0$ lies completely inside $x^2 +y ^2 +2b_2x+c=0$ then is
$(a) b_1b_2 > 0 $
$(b) b_1b_2 < 0$
$(c) c>0$
$(d)c<0$
Now I really couldn't find out how to solve this. Eg when I thought about $b_1$ or $b_2$, I imagined that since they are the centers of the 2 circle, they could have any sign, ie either both can be on the same side or on opposite sides of axis , but the correct choice is $b_1 b_2 >0$. For c, I don't even know where to begin with... Please help.
The answer given is (a) and (c)
Rewrite the two circles in their canonical forms,
$$(x-b_1)^2 +y^2 = b_1^2 -c $$
$$(x-b_2)^2 +y^2 = b_2^2 -c $$
The condition that the 1st circle lies completely inside the 2nd establishes the following inequality,
$$|b_2-b_1| + r_1 < r_2$$
or,
$$|b_2-b_1| + \sqrt{b_1^2-c} < \sqrt{b_2^2-c}$$
which means that the distance between the circles' centers should be smaller than the difference of their radii. Solve the inequality by squaring both sides
$$b_1(b_1-b_2)+|b_2-b_1|\sqrt{b_1^2-c}<0$$
Since the second term is positive, the following must hold,
$$b_1(b_1-b_2)<0$$
Case 1: $b_1<0$ and $b_1-b_2>0$, which leads to $0>b_1>b_2$;
Case 2: $b_1>0$ and $b_1-b_2<0$, which leads to $b_2>b_1>0$.
Therefore, the following holds in both cases,
$$b_1b_2>0$$
So, the answer is (a).