Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Find the equation of the chord of the circle  which is bisected at

Find the equation of the chord of the circle x^{2}+y^{2}=16 which is bisected at the point (-2,2) ?

Option: 1

x+y+4=0


Option: 2

x-y-4=0


Option: 3

x-2y+4=0


Option: 4

x-y+4=0


Answers (1)

best_answer

\\\text{The equation of the chord of the circle } x^{2}+y^{2}=16 \\\text{bisected at (-2,2) is } T=S_{1} \\\Rightarrow -2 x+2 y-16=4+4-16 \\ x-y+4=0

Posted by

Gaurav

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions