Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The locus of the foot of the perpendicular from the origin to the chords of the circle <img alt="\mathrm{x^2+y^2-2 x+4 y-1=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2

The locus of the foot of the perpendicular from the origin to the chords of the circle \mathrm{x^2+y^2-2 x+4 y-1=0} which subtend a right angle at the origin is
 

Option: 1

\mathrm{ x^2+y^2-4 x+8 y-2=0}


Option: 2

\mathrm{2 x^2+2 y^2-2 x+4 y-1=0}


Option: 3

\mathrm{ x^2+y^2-x+2 y-1=0}


Option: 4

\mathrm{ 2 x^2+2 y^2+2 x-4 y+1=0}


Answers (1)

best_answer

Let \mathrm{A B} is a chord which subtends 90^{\circ} at origin. Foot of perpendicular \mathrm{(h, k)} from origin is the mid point of chord \mathrm{A B}. Equation of chord \mathrm{A B} is

\begin{aligned} & \mathrm{y-k=-\frac{h}{k}(x-h) \Rightarrow k y+h x=h^2+k^2 }\\ & \mathrm{x^2+y^2-2 x\left(\frac{k y+h x}{h^2+k^2}\right)+4 y\left(\frac{k y+h x}{h^2+k^2}\right)-1\left(\frac{k y+h x}{h^2+k^2}\right)^2=0} \\ & \mathrm{\text { coeff of } x^2+\text { coeff of } y^2=0} \\ & \mathrm{2\left(h^2+k^2\right)-2 h+4 k-1=0} \end{aligned}

Posted by

shivangi.bhatnagar

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 session 2 answer key out for BArch, BPlanning; response sheet on jeemain.nta.nic.in
anam-khan

JEE Mains 2025 Paper 2 Answer Key LIVE: BArch final answer key PDF expected soon at jeemain.nta.nic.in
anam-khan

JEE Main 2025 response sheet errors: Delhi HC allows student to apply for JEE Adv, seeks NTA explanation

Latest Question