Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The equation of the circle having the lines $x^2+2 x y+3 x+6 y=0$ as its normals and having size just sufficient to contain the circle $x(x-4)+y(y-3)=0$ is
Option: 1
$x^2+y^2+3 x-6 y-40=0$

Option: 2
$x^2+y^2+6 x-3 y-45=0$

Option: 3
$x^2+y^2+8 x+4 y-20=0$

Option: 4
$x^2+y^2+4 x+8 y+20=0$

Answers (1)

best_answer

Given pair of normals is $x^2+2 x y+3 x+6 y=0 \, \, or\quad(x+2 y)(x+3)=0$

$\therefore \text { Normals are } x+2 y=0 \text { and } x+3=0$

The point of intersection of normals $x+2 y=0$ and $x+3=0$ is the centre

of required circle, we get centre $C_1=(-3,3 / 2)$ and other circle is

$x(x-4)+y(y-3)=0$ or $x^2+y^2-4 x-3 y=0$ $...(i)$

Its centre $C_2=(2,3 / 2)$ and radius $r=\sqrt{4+\frac{9}{4}}=\frac{5}{2}$

Since the required circle just contains the given circle (i), the given circle should touch the required circle internally from inside. Therefore, radius of the required circle $=\mid C_1-C_2 \downarrow+r$

$=\sqrt{(-3-2)^2+\left(\frac{3}{2}-\frac{3}{2}\right)^2}+\frac{5}{2}=5+\frac{5}{2}=\frac{15}{2}$
Hence, equation of required circle is

$(x+3)^2+\left(y-\frac{3}{2}\right)^2=\left(\frac{15}{2}\right)^2$

or $x^2+y^2+6 x-3 y-45=0$

Posted by

Rishabh

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow

anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern

anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question