Home
Class 12
MATHS
The equation of a circle C1 is x^2+y^2...

The equation of a circle `C_1` is `x^2+y^2= 4`. The locus of the intersection of orthogonal tangents to the circle is the curve `C_2` and the locus of the intersection of perpendicular tangents to the curve `C_2` is the curve `C_3`, Then

A

`C_(2)` is a circle

B

`C_(1), C_(2)` are circles having different centres

C

`C_(1), C_(2)` are circles having same centres

D

area enclosed between `C_(1) and C_(2)` is `8pi`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
Promotional Banner

Topper's Solved these Questions

  • CIRCLE

    VK JAISWAL|Exercise Exercise - 3 : Comprehension Type Problems|10 Videos
  • CIRCLE

    VK JAISWAL|Exercise Exercise - 4 : Matching Type Problems|2 Videos
  • CIRCLE

    VK JAISWAL|Exercise Exercise - 5 : Subjective Type Problems|13 Videos
  • BIONMIAL THEOREM

    VK JAISWAL|Exercise Exercise-4 : Subjective Type Problems|15 Videos
  • COMPLEX NUMBERS

    VK JAISWAL|Exercise EXERCISE-5 : SUBJECTIVE TYPE PROBLEMS|8 Videos

Similar Questions

Explore conceptually related problems

The equation of a circle is x^(2)+y^(2)+14x-4y+28=0. The locus of the point of intersection of orthoonal tangents to C_(1) is the curve C_(2) and the locus of the point of intersection of perpendicular tangents to C_(2) is the curve C_(3) then the statement(s) which hold good?

Locus of the point of intersection of perpendicular tangents to the circle x^(2)+y^(2)=16 is

Locus of the point of intersection of perpendicular tangents to the circle x^(2)+y^(2)=10 is

Locus of the point of intersection of perpendicular tangents to the circle x^(2)+y^(2)=16 is

Find the locus of the point of intersection of the perpendicular tangents of the curve y^(2)+4y-6x-2=0

The locus of the point of intersection of the perpendicular tangents to the parabola x^(2)=4ay is

Locus of point of intersection of perpendicular tangents to the circle x^(2)+y^(2)-4x-6y-1=0 is