Maximising the sum of squares of distance between 3 points on 3 concentric circles $\textbf{Question : }$ Given three concentric circles $C_1,C_2,C_3$ of radius $r_1,r_2,r_3\ (r_1<r_2<r_3)$ and a point on each of the circle's circumference then what is the maximum value of the sum of squares of the three lengths between the points?
In other words, calling $P_1,P_2,P_3$ the three points, maximise:
$$|P_1P_2|^2+|P_2P_3|^2+|P_3P_1|^2$$

$\textbf{My Approach}$
Let the complex number $|z_i|=r_i$ represent the circle $C_i$ in the complex plane. Then the quantity needed to be maximized will become
$$S=|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_1|^2$$
Opening this we get
$$S=2(|z_1|^2+|z_2|^2+|z_3|^2)-(z_1\overline{z_2}+\overline{z_1}z_2+z_2\overline{z_3}+\overline{z_2}z_3+z_3\overline{z_1}+\overline{z_3}z_1)$$
As we clearly know
$$|z_1+z_2+z_3|^2 \ge 0 \implies -(z_1\overline{z_2}+\overline{z_1}z_2+z_2\overline{z_3}+\overline{z_2}z_3+z_3\overline{z_1}+\overline{z_3}z_1) \le (|z_1|^2+|z_2|^2+|z_3|^2)$$
So
$$S_{max} = 3(|z_1|^2+|z_2|^2+|z_3|^2)$$
So I got the answer through this approach but I can't seem to do this using some geometrical interpretation of the above technique. I also did this : fix one point and take other two points as variables whose $x$-coordinates were $h,p$ and then found the general sum of squares of distances and differentiated wrt $h$ and equated to zero but that did not give me anything fruitful. So if anyone knows a geometrical method to this problem please give me some insight on how to do so...
 A: Fixing $z_1$ and $z_2$, the locus of $z_3$ for $L \stackrel{def}{=}|z_1-z_2|^2 + |z_2 - z_3|^2 + |z_3 - z_1|^2 = \text{constant}$ are circles centered at mid-point of $z_1,z_2$. For $z_3$ on the circle $|z| = r_3$ to maximize $L$, the locus circle need to tangent to circle $|z| = r_3$ at $z_3$. This implies the origin $0$ lies on the median of $\triangle z_1z_2 z_3$ (a line passing through $z_3$ and mid-point of $z_1,z_2$).
Apply same argument to other two pairs $z_2,z_3$ and $z_3,z_1$,
we find for the configuration where $L$ is maximized, the origin $0$ lies on the intersection of all three medians of $\triangle z_1z_2z_3$. If $\triangle z_1z_2z_3$ is non-degenerate (ie. $z_1$, $z_2$, $z_3$ are not collinear), the intersection is a single point and $0$ is centroid of $\triangle z_1z_2z_3$. In this case, $z_1 + z_2 + z_3 = 0$.
Update
To summarize, in order to maximize $L$, there are only two possible scenarios:

*

*$z_1, z_2, z_3$ are collinear. In this case, it is trivial to see
$L$ is maximized when $z_1,z_2$ is on opposite side of $z_3$ with respect to origin. A configuration that maximize $L$ is $(z_1,z_2,z_3) = (r_1,r_2,-r_3)$ with corresponding
$$\begin{align}
L = L_{1} &\stackrel{def}{=} (r_1+r_3)^2 + (r_2+r_3)^2 + (r_1-r_2)^2\\
&= 3(r_1^2+r_2^2+r_3^2) - (r_1+r_2-r_3)^2\end{align}
$$


*Otherwise, $z_1, z_2, z_3$ are not collinear and $z_1 + z_2 + z_3 = 0$ when $L$ is maximized. By a little bit of algebra, we have
$$L = L_{2} \stackrel{def}{=} 3(r_1^2+r_2^2+r_3^2)$$
The million dollar questions is

which scenarios give us the true maximum?

Notice $r_1 < r_2 < r_3$. When $r_3 > r_1 + r_2$. It is impossible to find $z_1, z_2, z_3$ on the circles to satisfy $z_1 + z_2 + z_3 = 0$. This rule out the $2^{nd}$ scenario and maximum $L$ is $L_1$.
On the other direction, let's say $r_3 < r_1 + r_2$. Among all configuration where $z_1, z_2, z_3$ is collinear, $(z_1,z_2,z_3) = (r_1,r_2,-r_3)$
remains to be a configuration that maximizes $L$.
Consider what happens if we keep $z_3$ fixed, increase the imaginary part of $z_1$ and $-z_2$ by a small $\epsilon$. To second order of $\epsilon$, the real part of $z_1$, $z_2$ will be decreased by amounts
$\frac{\epsilon^2}{2r_1}$ and $\frac{\epsilon^2}{2r_2}$ respectively. Plug this into expression of $L$. To second order of $\epsilon$ again, $L$ will be changed by an amount:
$$\begin{align}
\delta L &= 
\underbrace{2(r_3+r_1)\frac{-\epsilon^2}{2r_1}}_{\delta((x_1-x_3)^2)}
+
\underbrace{2(r_3+r_2)\frac{-\epsilon^2}{2r_2}}_{\delta((x_2-x_3)^2)}
+
\underbrace{2(r_2-r_1)\left(\frac{\epsilon^2}{r_1} - \frac{\epsilon^2}{r_2}\right)}_{\delta((x_1-x_2)^2)}
+ 6\epsilon^2 + o(\epsilon^2)\\
&= \frac{(r_1+r_2-r_3)(r_2+r_1)}{r_1r_2} \epsilon^2 + o(\epsilon^2)
\end{align}
$$
When $r_1 + r_2 > r_3$, the coefficient of $\epsilon^2$ in $\delta L$ is positive. This means the "maximum" from $1^{st}$ scenario cannot be the true maximum. The true maximum is $L_2$, the one from $2^{nd}$ scenario.
Combine all these, we have
$$\max L = 3(r_1^2+r_2^2+r_3^2) -
\begin{cases}(r_1 + r_2 - r_3)^2, & r_3 > r_1 + r_2\\0, &\text{ otherwise}
\end{cases}$$
