wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Locus of the point of intersection of perpendicular tangents to the circle x2+y2=16 is


A

x2+y2=8

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

x2+y2=32

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

x2+y2=64

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

x2+y2=16

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

The correct option is B

x2+y2=32


Explanation for the correct option:

Given the equation of the circle

x2+y2=16x2+y2=42

To find the locus of the point of intersection of perpendicular tangent.

We know that ,

Radius of locus = r2+r2=2r2

From the given equation radius is given as 4

Put r=4 in 2r2

Radius of locus =2×42=32

Therefore, the equation of the tangent

x2+y2=322x2+y2=32

Hence, option (B) is the correct answer.


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Line and Ellipse
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon