A circle of radius `r` passes through the origin `O` and cuts the axes at `A and B`. Let `P` be the foot of the perpendicular from the origin to the line `AB`. Find the equation of the locus of `P`.

A circle of radius 'r' passes through the origin O and cuts the axes at A and B, Locus of the centroid of triangle OAB is

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

A circle of constant radius r passes through the origin O, and cuts the axes at A and B. The locus of the foots the perpendicular from O to AB is (x^(2) + y^(2)) =4r^(2)x^(2)y^(2) , Then the value of k is

The locus of the foot of the perpendicular from the origin to a straight line passing through (1,1) is

A circle of radius 3k passes through (0.0) and cuts the axes in A and B then the locus of centroid of triangle OAB is

circle of radius 3k passes through (0,0) and cuts the axes in A and B then the locus of centroid of triangle OAB is

A circle of constant radius 2r passes through the origin and meets the axes in 'P' and 'Q' Locus of the centroid of the trianglePOQ is :

A circle of constant radius r passes through the origin O and cuts the axes at A and B. Show that the locus of the foot of the perpendicular from O to AB is (x^2+y^2)^2(x^(-2)+y^(-2))=4r^2 .