Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The centre of a circle passing through the points $(0,0),(1,0)$ and touching the circle $x^2+y^2=9$ is
Option: 1
$(3 / 2,1 / 2)$

Option: 2
$(1 / 3,3 / 2)$

Option: 3
$(1 / 2, \sqrt{2})$

Option: 4
$(1 / 2,-\sqrt{2})$

Answers (1)

best_answer

Let the equation of the circle be $x^2+y^2+2 g x+2 f y+c=0$ $...(i)$

It passes through the points $(0,0)$ and $(1,0),$

$\therefore \quad c=0$ and $1+2 g+c=0 ; \quad \therefore g=-1 / 2.$

Radius of circle (i) is $r_1=\sqrt{\left(g^2+f^2-c\right)}=\sqrt{\left(1 / 4+f^2\right)}$ .

The centre of the circle $x^2+y^2=9$ $...(ii)$

is $(0,0)$ and radius $r_2=3.$

Since the circle (i) passes through the centre $(0,0)$ of circle (ii) and it also touches the circle (ii) internally.

$\therefore \quad$ Diameter of circle (i) $=$ radius of circle (ii), i.e, $2 r_1=r_2$

or $2 \sqrt{\left(1 / 4+f^2\right)}=3 \Rightarrow f^2=2 \Rightarrow f= \pm \sqrt{2} .$

Hence the centre of circle (i) is $(-g,-f)$, i.e., $(1 / 2,-\sqrt{2})$ or $(1 / 2, \sqrt{2}).$

Hence (c) and (d) are correct answer.

Posted by

Ritika Harsh

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

Latest Question