If the circle $x^2+y^2+2ax+c=0$ lies completely inside the circle$ x^2+y^2+2 bx+c=0$, prove that$ ab>0,c>0$.

Question

I tried using the inequality, distance between centres is less than difference of the radiuses, but did not get any satisfactory result.Please help.

The distance between any point$(\sqrt{b^2-c}\cos t-b-a,\sqrt{b^2-c}\sin t)$ on $$x^2+y^2+2bx+c=0$$ from $(-a,0)$ should be greater than $$\sqrt{a^2-c}$$ — lab bhattacharjee, Aug 10 '16 at 05:05

score 6 · Accepted Answer · answered Aug 10 '16 at 05:08

6

If $c\le 0$, then $(0,\sqrt{-c})$ is on both circles, which is not allowed as the first lies completely inside the second. Therefore $c>0$.

We see that $(-a,0)$ and $(-b,0)$ are the centres of the circle. As the point $(0,0)$ is outside of both circles, we conclude first of all that $a\ne 0$ and $b\ne 0$, but also that both circles must be on the same side of the origin, which means $ab>0$.

answered Aug 10 '16 at 05:08

Hagen von Eitzen

374,180

Why should they be on the same side of the origin? – Areesh-Ul Eimaan Aug 10 '16 at 05:43

score 4 · Answer 2 · answered Aug 10 '16 at 08:21

4

The distance between the centers must be less than the radius of the larger circle. This gives the inequality $$(a-b)^2<b^2-c$$ which simplifies to $$a^2+c<2a b$$ It has been shown above that $c>0$ therefore $$a b>0$$

answered Aug 10 '16 at 08:21

Lozenges

1,742

If the circle $x^2+y^2+2ax+c=0$ lies completely inside the circle$ x^2+y^2+2 bx+c=0$, prove that$ ab>0,c>0$.

2 Answers2

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