Problem source

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 \textit{centroid} has coordinates $(\bar x, \bar y)$ where  $3\bar x = x_1+x_2+x_3,$ $3\bar y =y_1+y_2+y_3.
$]

Solution source

\begin{align*}
&& a^2 &= d_1^2 + d_2^2 + d_3^2 \\
&&&= (x-x_1)^2+(y-y_1)^2 + (x-x_2)^2+(y-y_2)^2 + (x-x_3)^2+(y-y_3)^2 \\
&&&= \sum (x-\bar{x}+\bar{x}-x_i)^2 + \sum (y-\bar{y}+\bar{y}-y_i)^2  \\
&&&= \sum \left ( (x-\bar{x})^2+(\bar{x}-x_i)^2 + 2(x-\bar{x})(\bar{x}-x_i) \right)+ \sum \left ( (y-\bar{y})^2+(\bar{y}-y_i)^2 + 2(y-\bar{y})(\bar{y}-y_i) \right)\\
&&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2 + 6x\bar{x} -6\bar{x}^2-2x\sum x_i+2\bar{x}\sum x_i + \\
&&&\quad\quad\quad 3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 + 6y\bar{y} -6\bar{y}^2-2y\sum y_i+2\bar{y}\sum y_i \\
&&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2+3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 \\
\\
\Rightarrow &&  (x-\bar{x})^2+(y-\bar{y})^2 &= \frac13\left ( a^2- \sum \left((\bar{x}-x_i)^2+(\bar{y}-y_i)^2 \right) \right)
\end{align*}

Therefore the locus is a circle, centre $(\bar{x}, \bar{y})$. radius $\sqrt{\frac13(a^2 - \text{sum of squares distances of centroid to vertices}})$. $a^2$ cannot be less than this distance, because clearly the right hand side is always bigger than it!