1.

The locus of the point, the sum of the squares of whose distances from n fixed points A_(i)(x_(i),y_(i))i=1,2,3....n. is equal to K^(2)is a circle.

Answer»

passing through the origin
with centre at origin
with centre at a point of mean position of the given POINTS
None of these

Solution :Let `(x,y)` be any point on the locus, then
`underset(i=1)overset(N)(Sigma)([x-x_(i)]^(2)+[y-y_(i)]^(2))=k^(2)impliesn(x^(2)+y^(2))-2x underset(i=1)overset(n)(Sigma)x_(i)-2yunderset(i=1)overset(n)(Sigma)y_(i)+underset(i=1)overset(n)(Sigma)x_(i)^(2)+underset(i=1)overset(n)(Sigma)y_(i)^(2)-K^(2)=0`
`implies x^(2)+y^(2)-2((1)/(n)underset(i=1)overset(n)(Sigma)x_(i))x-2((1)/(n)underset(i=1)overset(n)(Sigma)y_(i))y+(1)/(n)(underset(i=1)overset(n)(Sigma)x_(i)^(2)+underset(i=1)overset(n)(Sigma)y_(i)^(2)-K^(2))=0`
which is a CIRCLE with centre `((1)/(n)underset(i=1)overset(n)(Sigma)x_(i),(1)/(n)underset(i=1)overset(n)(Sigma)y_(i))` the point of mean position of the given points.


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