Home
Class 12
MATHS
A circle of constant radius a passes thr...

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`

Text Solution

Verified by Experts

The correct Answer is:
which is the required locus.
Promotional Banner

Topper's Solved these Questions

  • CIRCLE

    ARIHANT MATHS ENGLISH|Exercise Exercise For Session 1|18 Videos

  • CIRCLE

    ARIHANT MATHS ENGLISH|Exercise Exercise For Session 2|17 Videos

  • BIONOMIAL THEOREM

    ARIHANT MATHS ENGLISH|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|21 Videos

  • COMPLEX NUMBERS

    ARIHANT MATHS ENGLISH|Exercise Complex Number Exercise 8|2 Videos

Similar Questions

Explore conceptually related problems

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

A circle of radius r passes through origin and cut the x-axis and 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^2y^2) (B) (x^2+y^2)^2(x+y)=r^2(xy) (C) (x^2+y^2)^2=r^2(x^2y^2) (D) (x^2+y^2)^3=4r^2(x^2y^2)

The locus of the foot of the perpendicular from the center of the hyperbola x y=1 on a variable tangent is (a) (x^(2)+y^(2))^(2)=4xy (b) (x^2-y^2)=1/9 (c) (x^2-y^2)=7/(144) (d) (x^2-y^2)=1/(16)

The locus of the foot of the perpendicular from the center of the hyperbola x y=1 on a variable tangent is (x^2+y^2)^2=4x y (b) (x^2-y^2)=1/9 (x^2-y^2)=7/(144) (d) (x^2-y^2)=1/(16)

The locus of the foot of the perpendicular from the center of the hyperbola x y=1 on a variable tangent is (a) (x^2+y^2)^2=4x y (b) (x^2-y^2)=1/9 (x^2-y^2)=7/(144) (d) (x^2-y^2)=1/(16)

The locus of the foot of perpendicular drawn from the centre of the ellipse x^2+""3y^2=""6 on any tangent to it is (1) (x^2-y^2)^2=""6x^2+""2y^2 (2) (x^2-y^2)^2=""6x^2-2y^2 (3) (x^2+y^2)^2=""6x^2+""2y^2 (4) (x^2+y^2)^2=""6x^2-2y^2

The equation of the curve passing through origin, whose slope at any point is (x(1+y))/(1+x^2) , is (A) (1+y)^2-x^2=1 (B) x^2+(y+1)^2=1 (C) (x+y)y=1-x^2 (D) x=ye^((1+y))

If tangents are drawn to the ellipse x^2+2y^2=2, then the locus of the midpoint of the intercept made by the tangents between the coordinate axes is (a) 1/(2x^2)+1/(4y^2)=1 (b) 1/(4x^2)+1/(2y^2)=1 (c) (x^2)/2+y^2=1 (d) (x^2)/4+(y^2)/2=1

The locus of the foot of the perpendicular from the origin on each member of the family (4a+ 3)x - (a+ 1)y -(2a+1)=0 (a) (2x-1)^(2) +4 (y+1)^(2) = 5 (b) (2x-1)^(2) +(y+1)^(2) = 5 (c) (2x+1)^(2)+4(y-1)^(2) = 5 (d) (2x-1)^(2) +4(y-1)^(2) = 5

The radius of the circle passing through the points of intersection of ellipse x^2/a^2+y^2/b^2=1 and x^2-y^2 = 0 is