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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)

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Mr.HiggsBoson
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1 Answers1

1

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).

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