A circle of radius r passes through the origin and intersects the x-axis and y-axis at P and Q respectively. Show that the equation to the locus of the foot of the perpendicular drawn form the origin upon the line segment `bar (PQ)` is `(x^(2)+y^(2))^(3)=4r^(2)x^(2)y^(2)`.

A circle of constant radius 'r' passes through origin O and cuts the axes of coordinates in points P and Q, then the equation of the locus of the foot of perpendicular from O to PQ is

A line passing through the point P(4,2) meets the x and y-axis at A and B respectively. If O is the origin, then locus of the centre of the circumcircle of triangle OAB is

Find the locus of the foot of the perpendicular from the origin upon a stright line which always passes through the point (2,3).

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

Find the equation of a circle which passes through the origin and makes intercepts of 2 units and 4 units on x-axis and y-axis respectively.

A plane passes through a fixed point (a ,b ,c)dot Show that the locus of the foot of the perpendicular to it from the origin is the sphere x^2+y^2+z^2-a x-b y-c z=0.

Radius of the circle that passes through the origin and touches the parabola y^2=4a x at the point (a ,2a) is

A striaght line intersect x-axis at A(7,0) and y-axis at B(0,-5). The straight line PQ is perpendicular to AB and intersect the x and y-axes at P adn Q respectively. Find the eqution to the locus of the point of intersection of the line AQ and BP.

A circle passes through the origin and has its center on y=x If it cuts x^2+y^2-4x-6y+10=0 orthogonally, then find the equation of the circle.

Line segment AB of fixed length c slides between coordinate axes such that its ends A and B lie on the axes. If O is origin and rectangle OAPB is completed, then show that the locus of the foot of the perpendicular drawn from P to AB is x^((2)/(3)) + y^((2)/(3)) = c^((2)/(3)).