A variable circle c, having a fixed radius r passes through the origin and cuts the axes in A, B. OP is perpendicular to AB Prove that P traces the curve `(x^2+y^2)^3=4r^2x^2y^2`

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

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))^k =4r^(2)x^(2)y^(2) , Then the value of k 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 .

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 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 . 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 origin and cut the y-axis at P and Q. The locus of foot of perpendicular drawn from origin upon line joining the points P and Q is (A) (x^(2)+y^(2))^(3)=r^(2)(x^(2)y^(2))(B)(x^(2)+y^(2))^(2)(x+y)=r^(2)(xy)(C)(x^(2)+y^(2))^(2)=r^(2)(x^(2)y^(2))(D)(x^(2)+y^(2))^(3)=4r^(2)(x^(2)y^(2))

If the radius of the circle passing through the origin and touching the line x+y=2 at (1, 1) is r units, then the value of 3sqrt2r 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 (A) (x^2+y^2)(1/(x^2)+1/(y^2))=4a^2 (B) (x^2+y^2)^2(1/(x^2)+1/(y^2))=a^2 (C) (x^2+y^2)^2(1/(x^2)+1/(y^2))=4a^2 (D) (x^2+y^2)(1/(x^2)+1/(y^2))=a^2