Question: A point moves in the x-y plane so that the sum of the squares of its distances from the three fixed points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is always $a^2$ . Find the equation of the locus of the point and interpret it geometrically. Explain why $a^2$ cannot be less than the sum of the squares of the distances of the three points from their centroid.
The locus of the points can be found by just using letting the point be $(x,y)$ and using the distance formula.
I am new to this format so excuse me for doing this:
Let $S_1$=$(x_1+x_2+x_3)$, $S_2$=$((x_1)^2+(x_2)^2+(x_3)^2)$, $S_3$=$(x_1x_2+x_1x_3+x_2x_3)$
Also let $S_4$,$S_5$,$S_6$ be similar expressions but with $y$
After some tedious algebra you can say that $$(x-(S_1)/3)^2+(y-(S_4)/3)^2=a^2/3-2/9(S_2+S_5-S_3-S_6)$$ Hence this is a circle centred about the centroid.
Now here is where my problem lies, only an unofficial solution exists and it's at this point that they jump to the conclusion that if $a^2$ is less than the sum of distances from the centroid to each of the points then the radius would have to be negative hence that would be impossible.
I cannot understand how they came to that conclusion so if someone could help me finish this line of reasoning I would be grateful.
I did come up with my own solution but I'd like to understand the other solution as well, here's my solution if anyone's interested:
The locus can also be found by drawing 3 circles with radii $r_1$,$r_2$,$r_3$ each centred about the points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ where $r_1^2$+$r_2^2$+$r_3^2$=$a^2$.
You can justify this by looking at the equation immediately after applying the distance formula to each of the points.
Now, every point in the locus is a point where the 3 circles intersect in any possible configuration of the radii.
The minimum of the sum of squares of the radii occurs when all the radii are equal by the RMS-AM-GM inequality and this only occurs at the centroid hence the minimum of $a^2$ is the sum of the squares of the distances between the centroid and the points.
Therefore, $a^2$ cannot be less than this sum.
Let $\bar{x} = {1 \over 3} (x_1+x_2+x_3)$.
Then $\sum_k \|x-x_k\|^2 = 3 \|x\|^2 + \sum_k \|x_k\|^2 - 3 \cdot 2 x^T \bar{x}$ and
$\sum_k \| x_k -\bar{x}\|^2 = 3 \|\bar{x}\|^2 + \sum_k \|x_k\|^2 - 3 \cdot 2\|\bar{x}\|^2 = \sum_k \|x_k\|^2 - 3\|\bar{x}\|^2$.
Hence \begin{eqnarray} \sum_k \|x-x_k\|^2 &=& 3 \|x\|^2+ \sum_k \| x_k -\bar{x}\|^2 + 3\|\bar{x}\|^2- 3 \cdot 2 x^T \bar{x} \\ &=& 3 \|x-\bar{x}\|^2+\sum_k \| x_k -\bar{x}\|^2 \end{eqnarray} and so the equation reduces to $3 \|x-\bar{x}\|^2 = a^2-\sum_k \| x_k -\bar{x}\|^2$