Prove that the circles are non-intersecting in a coaxial system of circles $x^2+y^2+2gx+c=0$

Question

In a coaxial system of circles $x^2+y^2+2gx+c=0$ where $g$ is a parameter, if $c>0$ then prove that the circles are non-intersecting

Let pair of circles be $x^2+y^2+2g_1x+c=0$ and $x^2+y^2+2g_2x+c=0$

The radial axis is $$ (g_1-g_2)x=0\implies x=0 $$ Center $(-g,0)$

Radius =$\sqrt{g^2-c}$

Distance between radial axis and the center $=g_1$

I think if it is non-intersecting $g_1>\sqrt{g_1^2-c}$ but I don't think there is enough information to conclude that ?

Answer 1

Obviously, $|g| > \sqrt{c}$, otherwise there is real solutions for $x^2+y^2 + 2gx + c =0$.

---

When $g_1\cdot g_2<0$, or in other words, $g_2\geq \sqrt{c} > -\sqrt{c} >g_1$,

we have $g_1 + \sqrt{g_1^2-c} < 0 < g_2 - \sqrt{g_2^2-c}$. The 2 circles are in 2 side of the $y$ axis. Thus, not intersecting

---

when $g_1\cdot g_2>0$. We can assume $|g_2| > |g_1| \geq \sqrt{c}$ without the loss of generosity. $$ \sqrt{g_2^2 -c} - \sqrt{g_1^2-c} = \frac{g_2^2- g_1^2}{\sqrt{g_2^2 -c} + \sqrt{g_1^2-c}} > \frac{|g_2|^2- |g_1|^2}{\sqrt{g_2^2} + \sqrt{g_1^2}} = \frac{|g_2|^2- |g_1|^2}{|g_2| + |g_1|} = |g_2| - |g_1| $$

The difference in their radius is larger than the distance in their center. Thus, still not intersecting.

---

In summary, they are not intersecting in all situations.

Answer 2

The equation of circle $C_g$ can be written :

$$(x+g)^2+y^2=R^2 \ \ \text{with} \ \ R=\sqrt{g^2-c}$$

(in fact, we must take $|g|>\sqrt{c}$). Let $D_g$ be the disk associated with $C_g$ (its interior set).

Fig. 1 : (here $c=2$). The case $g>0$ (resp $g<0$) corresponds to the red (resp. blue) circles situated in halfplane $x<0$, (resp. $x>0$).

Let us consider in the following lines the case : $g>0$ corresponding to red circles in the figure (center $(-g,0)$ and radius $R=\sqrt{g^2-c}$ with $R<g$). Disk $D_g$, being the set of points that are at a distance less that $R<g$ of $(g,0)$ is entirely included in the left hand side half plane, i.e., with $x<0$.

As we have $x<0$ for all points in the interior of any circle $C_g$, under the hypothesis $g_1>g_2$, we have, for any $x<0$ : $2g_1x<2g_2x$.

Therefore, adding $x^2+y^2+c$ to the LHS and RHS of the previous identity, we preserve the inequality :

$$\underbrace{x^2+y^2+2g_1x+c<0}_{\text{Disk} D_{g_1}} \ \implies \ \underbrace{x^2+y^2+2g_2x+c<0}_{\text{Disk} D_{g_2} }$$

As implication of properties is to be interpreted as inclusion of sets, we have

$$g_1>g_2 \implies D_{g_1} \subset D_{g_2}\tag{1}$$

These disks are thus included like Russian dolls. It remains to prove that we have a strict inclusion in (1) ; let us assume on the contrary that we have a common point (at the boundary) i.e., there exists $(x,y)$ such that we have simultaneously :

$$\begin{cases}x^2+y^2+2g_1x+c=0\\x^2+y^2+2g_2x+c=0\end{cases}$$

Then we would have by difference, $(g_1-g_2)x=0$, thus $x=0$ which is impossible for this left (blue) family of circles.

The second case : $g<0$ (in fact $g<-\sqrt{c}$) is symmetrical of the first case (blue circles for which all points have a positive abscissa).

Final argument : none of the (blue) circles of the second family intersect the (red) circles of the first family because of this segregation by abscissa $(x<0/x>0)$.