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  • If the locus of a point, the sum of the squares of whose distances from n fixed points  <img alt="\mathrm{A_i\left(x_i, y_i\right), i=1,2}" src="https://en

If the locus of a point, the sum of the squares of whose distances from n fixed points 

\mathrm{A_i\left(x_i, y_i\right), i=1,2}, ... n is equal to \mathrm{k^{2}} is the circle \mathrm{x^2+y^2+2 g x+2 f y+c=0}, then 

 

Option: 1

\mathrm{g=-\frac{1}{n} \sum x_i y_i}


Option: 2

\mathrm{f=-\frac{1}{n} \sum x_i y_i}


Option: 3

\mathrm{c=k^2}


Option: 4

\mathrm{c=\frac{1}{n}\left(\sum x_i^2+\sum y_i^2-k^2\right)}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \text { Locus is } \sum_{i=1}^n\left[\left(x-x_i\right)^2+\left(y-y_i\right)^2\right]=k^2 \\ & =x^2+y^2-\frac{2 \sum x_i x}{n}-\frac{2 \sum y_i y}{n}+\frac{\sum x_i^2+\sum y_i^2-k^2}{n}=0 \\ & \Rightarrow c=\frac{1}{n}\left(\sum x_i^2+\sum y_i^2-k^2\right) \end{aligned}}

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