Question

The locus of the point, the sum of the squares of whose distances from n fixed points $A(x_1,y_1),i=1,2,...n$ is equal to $k^2$ is a circle

• A. passing through the origin
• B. with centre at the origin
• C. with centre at the point of mean position of the given points
• D. None of these

Answer 1

Answer: (C)

with centre at the point of mean position of the given points

Let $(x,y)$ be any point on the locus, then

$\sum _{i=1}^n[{(x-x_i)}^2+{(y-y_i)}^2]=k^2$

$\Rightarrow n(x^2+y^2)-2x\sum _{i=1}^n{x}_i-2y\sum _{i=1}^n{y}_i+\sum _{i=1}^n{x_i}^2+\sum _{i=1}^n{y_i}^2-k^2=0$

$\Rightarrow x^2+y^2-2\left(\frac{1}{n}\sum _{i=0}^n{x}_i\right)x-2\left(\frac{1}{n}\sum _{i=1}^n{y}_i\right)y+\frac{1}{n}(\sum _{i=1}^n{x_i}^2+\sum _{i=1}^n{y_i}^2-k^2)=0$

which is $a$ circle with center $(\frac{1}{n}\sum _{i=1}^n{x}_i,\frac{1}{n}\sum _{i=1}^n{y}_i)$ the point of mean position of the given points.