Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The centres of the circle passing through the point (0,0), (1,0) and touching the circle <img alt="\mathrm{ x^2+y^2=9}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%2

The centres of the circle passing through the point (0,0), (1,0) and touching the circle \mathrm{ x^2+y^2=9} are,

Option: 1

\left(\frac{1}{2}, \pm 2\right)


Option: 2

\left(\frac{1}{2}, \pm \sqrt{2}\right)


Option: 3

(1, \pm \sqrt{2})


Option: 4

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


Answers (1)

best_answer

Circle through (0,0) and (1,0) is

\mathrm{ \begin{aligned} & x^2+y^2-x+2 f y=0 \quad(\because c=0,2 g=-1) \\ & C_1(1 / 2,-f), r_1=\sqrt{\left(\frac{1}{4}+f^2\right)} \end{aligned} }

and \mathrm{C_2(0,0), r_2=3.}

If the two circles touch, then

\mathrm{ \begin{aligned} & C_1 C_2=r_2-r_1 \text { or } r_2+r_1 \\\\ \therefore & \sqrt{\left(\frac{1}{4}+f^2\right)}=3 \pm \sqrt{\left(\frac{1}{4}+f^2\right)} \\\\ \Rightarrow & \sqrt{\left(\frac{1}{4}+f^2\right)}=\frac{3}{2} \text { or } 0=3 \text { (rejected) } \\\\ \Rightarrow & f^2+1 / 4=9 / 4 \Rightarrow f= \pm \sqrt{2} \\\\ \therefore & \text { Centres are }(1 / 2, \pm \sqrt{2}) \end{aligned} }

Posted by

vinayak

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question