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`.

A circle of constant radius 5 units passes through the origin O and cuts the axes at A and B. Then the locus of the foot of the perpendicular from O to AB is (x^(2)+y^(2))^(2)(x^(-1)+y^(-2))=k then k=

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

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 a passes through the origin O and cuts the axes of coordinates at points P and Q . Then the equation of the locus of the foot of perpendicular from O to P Q is (x^2+y^2)(1/(x^2)+1/(y^2))=4a^2 (x^2+y^2)^2(1/(x^2)+1/(y^2))=a^2 (x^2+y^2)^2(1/(x^2)+1/(y^2))=4a^2 (x^2+y^2)(1/(x^2)+1/(y^2))=a^2

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

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 :

If a circle of constant radius 3k passes through the origin O and meets the coordinate axes at AandB,then the locus of the centroud of triangle OAB is (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)

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 .