If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is

Options

A. `x^(2)+y^(2)=(2k)^(2)`
B. `x^(2)+y^(2)=(3k)^(2)`
C. `x^(2)+y^(2)=(4k)^(2)`
D. `x^(2)+y^(2)=(6k)^(2)`

Answer

1

Explanation

Let the centroid of triangle OAB be (p,q).
Hence, points A and B are (3pm,0) and (0,3q), respectively.

But diameter of circle, `AB=6k`
Hence, `sqrt(9p^(2)+9q^(2))=6k`
Therefore, the locus of (p,q) is `x^(2)+y^(2)=4k^(2)`