Question

A point moves so that the sum of the squares of its distances from $n$ fixed points is given. Prove that its locus is a circle.

Answer 1

Let the points be $(x_1,y_1),(x_2,y_2),(x_3,y_3)....(x_n,y_n)$ and let the given sum of distances be t

The point is given by (x,y)

$(x-x_1{)}^2+(y-y_1{)}^2+(x-x_2{)}^2+(y-y_2{)}^2+(x-x_3{)}^2+(y-y_3{)}^2+...=t$

$\Rightarrow n{x}^2+n{y}^2-2(x_1+x_2+x_3+...x_n)x-2(y_1+y_2+y_3+...y_n)y+$

$(x_1^2+x_2^2+x_3^2+...)+(y_1^2+y_2^2+y_3^2...)=t$

$\Rightarrow x^2+y^2-\frac{2}{n}(x_1+x_2+x_3+....)-\frac{2}{n}(y_1+y_2+y_3+....)+\frac{(x_1+x_2+x_3+...{)}^2}{n^2}+\frac{(y_1+y_2+y_3+...{)}^2}{n^2}=\frac{t}{n}-\frac{(x_1^2+x_2^2+x_3^2+...)}{n}-\frac{(y_1^2+y_2^2+y_3^2+...)}{n}+\frac{(x_1+x_2+x_3+...{)}^2}{n^2}+\frac{(y_1+y_2+y_3...{)}^2}{n^2}$

$\Rightarrow {(x-\frac{(x_1+x_2+x_3+...)}{n})}^2+{(y-\frac{(y_1+y_2+y_3+...)}{n})}^2=\frac{t}{n}-\frac{(x_1^2+x_2^2+...)}{n}-\frac{(y_1^2+y_2^2+..)}{n}+\frac{(x_1+x_2+...{)}^2}{n^2}+\frac{(y_1+y_2+...{)}^2}{n^2}$

This is an equation of circle with centre at $(\frac{x_1+x_2+x_3+...}{n},\frac{y_1+y_2+y_3+...}{n})$

Hence proved, the locus of the point is a circle.