A point `P` moves in such a way that the ratio of its distance from two coplanar points is always a fixed number `(!=1)` . Then, identify the locus of the point.

A point moves in the xy-plane such that the sum of its distance from two mutually perpendicular lines is always equal to 3. The area of the locus of the point is

A point moves in the XY-plane such that the sum of it’s distances from two mutually perpendicular lines is always equal to 5 units. The area enclosed by the locus of the point is in square units is x then sum of the digits of x is

The locus of the point which moves such that the ratio of its distance from two fixed point in the plane is always a constant k(< 1) is

A point P moves such that its distances from two given points A and B are equal. Then what is the locus of the point P ?

A point moves in the x-y plane such that the sum of its distances from two mutually perpendicular lines is always equal to sqrt(2) units.The area enclosed by the locus of the point is (in sq.units)

A point moves so that the sum of squares of its distances from the points (1,2) and (-2,1) is always 6. then its locus is

Show that the locus of a point such that the ratio of its distances from two given points is constant,is a circle.Hence show that the circle cannot pass through the given points.