Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The centre of circle  lies on <img alt="\mathrm{ 2 x-2 y+9=0}" src="https://entrancecorner.on

The centre of circle \mathrm{ S} lies on \mathrm{ 2 x-2 y+9=0} and it cuts orthogonally the circle \mathrm{x^2+y^2=4}. Then the circle passes through two fixed points
 

Option: 1

(1,1),(3,3)

 


Option: 2

\left(-\frac{1}{2}, \frac{1}{2}\right),(-4,4)
 


Option: 3

(0,0),(5,5)
 


Option: 4

none of these


Answers (1)

best_answer

Let \mathrm{S=x^2+y^2+2 g x+2 f y+c=0}

\mathrm{\triangle\: it \: cuts \: x^2+y^2=4}orthogonally

\mathrm{ \therefore \quad 2 g_1 g_2+2 f_1 f_2=c_1+c_2 }

\mathrm{\Rightarrow \quad c=4 }

\mathrm{ (-g,-f) \text { lies on } 2 x-2 y+9=0 }

\mathrm{ \therefore-2 g+2 f+9=0 }

\mathrm{ \therefore \quad S \equiv x^2+y^2+2 g x+2 f y+4=0 }

\mathrm{\Rightarrow \quad x^2+y^2+(2 f+9) x+2 f y+4=0 \Rightarrow\left(x^2+y^2+9 x+4\right)+2 f(x+y)=0}
It is of the form \mathrm{S+\lambda P=0} and hence passes through the intersection of \mathrm{S=0 \: and\: P=0} intersection point are \mathrm{\left(-\frac{1}{2}, \frac{1}{2}\right),(-4,4).}

Hence option 2 is correct.

Posted by

Devendra Khairwa

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 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question