CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The equation of the circle through the points of intersection of x2+y21=0,x2+y22x4y+1=0 and touching the line x + 2y = 0, is


A

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C


Family of circles is

x2+y22x4y+1+λ(x2+y21)=0.

(1+λ)x2+(1+λ)y22x4y+(1λ)=0

x2+y221+λx41+λy+1λ1+λ=0 ............(i)

Centre is [11+λ,21+λ]

and radius = (11+λ)2+(21+λ)21λ1+λ=4+λ21+λ.

Since it touches the line x + 2y = 0, hence

Radius = Perpendicular from centre to the line

i.e., 11+λ+221+λ12+22 = 4+λ21+λ

5=4+λ2=±1

λ=1 cannot be possible in case of circle. So λ=1.

Thus, from (i) x2+y2x2y=0 is the required equation of the circle.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Family of Circles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon