Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The tangent to the circle C_{1}:x^{2}+y^{2}-2x-1=0 at the point (2, 1) cuts off a chord of length 4 from a circle C_{2} whose centre is (3, −2). The radius of C_{2} is :

  • Option 1)

    2

  • Option 2)

    \sqrt{2}

  • Option 3)

    3

  • Option 4)

    \sqrt{6}

 

Answers (1)

best_answer

As we learned,

 

General form of a circle -

x^{2}+y^{2}+2gx+2fy+c= 0
 

- wherein

centre = \left ( -g,-f \right )

radius = \sqrt{g^{2}+f^{2}-c}

 

 and

 

Equation of tangent -

xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0
 

- wherein

Tangent to circle

x^{2}+y^{2}+2gx+2fy+c=0  at  (x_{1},y_{1})

 

 

 Equation of tangent on C_{1} at (2,1) is 

2x+y-(x+2)-1=0

\Rightarrow x + y = 3

It cuts off circle C_{2} ;

distance from centre (3,-2)

\Rightarrow \: \left | \frac{3-2-3}{\sqrt{2}} \right |= \sqrt{2}

Length of chord = 4

r^{2}=\frac{l^{2}}{4+d^{2}}\: \Rightarrow \: r^{2}=4+2=6

r=\sqrt{6}


Option 1)

2

Option 2)

\sqrt{2}

Option 3)

3

Option 4)

\sqrt{6}

Posted by

Himanshu

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 2024 session 2 result date announced; Steps to check NTA session 2 scores
anam-khan

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today

Latest Question