Circle containing other circle Below is the question I am referring to:

Two circles have the equations $x^2+y^2+\lambda x +c=0$ and $x^2+y^2+\mu x + c = 0$. Prove that one of the circles will be within the other if $\lambda\mu>0$ and $c>0$.

What I did:For one circle to contain the other circle,the distance between their centres should be less than absolute value of difference between their radii.So I tried to prove this inequality with the given constraints that $\lambda\mu >0   ,c>0$.So,is this way to do the problem justified and what could be other ways to do the problem.Below are the images of my work.Thank you.
Part 1 of work:

Part 2 of work:

 A: The circle equations have no coefficient for $y^1$, and also putting $ y=-y$ there is no change. We can  delete $y^2$ term to see ordering of roots on x-axis.So both circles are centered on x-axis.
Product of roots (power of circles = $c$ ) is same. Center distances are half sum of roots.$ -\lambda/2, - \mu/2 $ and product of roots $c>0$.
If $\mu$ circle are to be inside $\lambda $ circle,
the  roots need to be ordered in positive x-direction left to right( where $B$ is for Big circle and $S$ for small circle ) in the following manner: 
$$ x^2 + \lambda x + c =0,\, x^2 + \mu x + c =0,\, $$
$$ B1= \frac{ -\lambda - \sqrt { \lambda^2-4c} }{2} ,S1=\frac{ -\mu - \sqrt { \mu^2-4c} }{2} ,$$
$$ S2=\frac{  -\mu + \sqrt { \mu^2-4c} }{2},B2=\frac{- \lambda + \sqrt { \lambda^2-4c} }{2} ; $$
as also shown in  figure.

Algebraic criterion for such geometric ordering is :
$$ (S1-B1)( B2-S2) >0$$
or
$$ (S1 B2+S2 B1) - (S1 S2+B1 B2) >0 $$
Simplfying (omitting intermediate steps),
$$ \lambda \mu >  \sqrt{\lambda^2-4 c}\, \sqrt {\mu^2 - 4 c} +4 c $$
The quantity under radical sign is positive discriminant of quadratic equation because there are two distinct real roots for each $\lambda$ and $\mu$ circles as given above.
$ c>0$ for positive power of circles shown, so  
$$ \lambda \mu > 0. $$
This means that wlog both circles can be reflected about $y$axis.
A: The circle equations are of the form
$$
0 = x^2 + y^2 + tx + c = (x + t/2)^2 + y^2 + c - t^2/4 \iff \\
\left( x + \frac{t}{2} \right)^2 + y^2 = \left( \frac{\sqrt{t^2 - 4c}}{2} \right)^2 = r^2
$$
which indeed is a circle with center $(x_0, y_0) = (-t/2, 0)$ and radius 
$$
r = \frac{\sqrt{t^2 - 4c}}{2} = \sqrt{x_0^2 - c}
$$ 
under the condition $ t^2 - 4c \ge 0$ or $c \le t^2/4 = x_0^2$.
We have $c>0$ thus $r < \sqrt{x_0^2} = \lvert x_0 \rvert$, 
which means the whole circle (with $x$ coordinates ranging from $x_0 - r$ to $x_0 + r$) is either on the half plane $x < 0$ or on the half plane $x > 0$.
We have $\lambda \mu > 0$ so either both $\lambda$ and $\mu$ are positive or both are negative.
This means both circles are either on the half plane $x < 0$ or half plane $x > 0$.
So we have circles for $\lvert x_0 \rvert \ge \sqrt{c}$, which then have radius $r(\lvert x_0 \rvert) = \sqrt{\lvert x_0 \rvert^2 - c}$ this means the larger $\lvert x_0 \rvert$, the larger the radius is.
If $\lvert \lambda \rvert \le \lvert \mu \rvert$ then 
$\lvert x_0^{(\lambda)} \rvert \le \lvert x_0^{(\mu)} \rvert$
and the first circle $C_\lambda$ will have a smaller or equal radius $r_\lambda$ than the second circle $C_\mu$ with radius $r_\mu$.
The points $(x,y)$ on the first circle $C_\lambda$ fulfill
$$
\left(x - x_0^{(\lambda)}\right)^2 + y^2 =
r_\lambda^2 \iff \\
\left(x - x_0^{(\lambda)}\right)^2 + y^2 = 
\left( x_0^{(\lambda)} \right)^2  - c \iff \\
x^2 - 2 x x_0^{(\lambda)} + y^2 = -c \\
$$
We now transform the left hand side into the left hand side of the equation for the second circle:
$$
x^2 + y^2 = -c + 2 x x_0^{(\lambda)} \iff \\
x^2 - 2 x x_0^{(\mu)} + \left( x_0^{(\mu)} \right)^2 + y^2 = 
\left( x_0^{(\mu)} \right)^2 - c + 
2 x x_0^{(\lambda)} - 2 x x_0^{(\mu)} \iff \\
\left( x - x_0^{(\mu)} \right)^2 + y^2 = 
\left( x_0^{(\mu)} \right)^2 - c + 
2 x \left( x_0^{(\lambda)} - x_0^{(\mu)} \right) 
= r_\mu^2 + \Delta 
$$
with
$\Delta = 2 x \left( x_0^{(\lambda)} - x_0^{(\mu)} \right)$.
For circles on the half plane $x > 0$ we have
$x_0^{(\lambda)} < x_0^{(\mu)}$ and $\Delta$ is negative.
For circles on the half plane $x < 0$ we have
$x_0^{(\lambda)} > x_0^{(\mu)}$ and $\Delta$ is negative too.
This means for all points on the first circle we have
$$
\left( x - x_0^{(\mu)} \right)^2 + y^2 = 
r_\mu^2 + \Delta = r(x,y)^2 < r_\mu^2
$$
in other words: The first circle lies within the second circle.
I leave the case $\lvert \mu \rvert \le \lvert \lambda \rvert$ to the reader.
A: $$d < R-r$$
$$\left| \frac{\lambda-\mu}{2} \right| <
  \left| \sqrt{\frac{\lambda^2}{4}-c}-\sqrt{\frac{\mu^2}{4}-c} \right| $$
$$\left( \frac{\lambda-\mu}{2} \right)^2 <
  \left( \sqrt{\frac{\lambda^2}{4}-c}-\sqrt{\frac{\mu^2}{4}-c} \right)^2$$
$$\frac{\lambda^2}{4}-\frac{\lambda \mu}{2}+\frac{\mu^2}{4} <
  \frac{\lambda^2}{4}+\frac{\mu^2}{4}-2c-
 2\left( \sqrt{\frac{\lambda^2}{4}-c} \, \right)
  \left( \sqrt{\frac{\mu^2}{4}-c} \, \right) $$
For real circles,
$$\lambda^2 > 4c \quad \text{and} \quad \mu^2 > 4c$$
Therefore

$$0<
  \left( \sqrt{\frac{\lambda^2}{4}-c} \, \right)
  \left( \sqrt{\frac{\mu^2}{4}-c} \, \right)
  < \frac{\lambda \mu}{4}-c \quad \cdots \cdots (*) $$

$$\left( \frac{\lambda^2}{4}-c \right)
  \left( \frac{\mu^2}{4}-c \right)
< \left( \frac{\lambda \mu}{4}-c \right)^2 $$
$$-\frac{c(\lambda^2+\mu^2)}{4} < -\frac{c\lambda \mu}{2}$$
$$0 < c (\lambda-\mu)^2$$
For $\lambda \neq \mu$,

$$c>0$$

But from $(*)$, $$\lambda \mu>4c$$
Hence

$$\lambda \mu >0$$

