Get Answers to all your Questions

header-bg qa

Tangents are drawn to the circle \mathrm{x}^2+\mathrm{y}^2=1  at the point where it is met by the circles \mathrm{x^2+y^2+(\lambda+2) x+(4-2 \lambda) y+1=0, \lambda}  being the parameter. The locus of the point of intersection of these tangents is
 

Option: 1

\mathrm{ 2 x+3 y+1=0}


Option: 2

\mathrm{ x+2 y+4=0}


Option: 3

\mathrm{ 2 x+y+4=0}


Option: 4

\mathrm{x}-2 \mathrm{y}-1=0


Answers (1)

best_answer

let \left(\mathrm{x}_1, \mathrm{y}_1\right)  be the point of intersection of tangents then eq. of chord of contact of \left(\mathrm{x}_1, \mathrm{y}_1\right)\text{ is }\mathrm{xx}_1+ \mathrm{yy}_1-1=0\quad \quad \cdots(1)  also common chord of two circles is \mathrm{S}-\mathrm{S}^1=0

\Rightarrow(\lambda+2) \mathrm{x}+(4-2 \lambda) \mathrm{y}+2=0\quad \quad \dots(2)

Eq. (1) & (2) represents the same line

\frac{\lambda+2}{\mathrm{x}_1}=\frac{4-2 \lambda}{\mathrm{y}_1}=\frac{2}{-1} \Rightarrow 2 \mathrm{x}_1+\mathrm{y}_1+4=0
 

Posted by

manish

View full answer

JEE Main high-scoring chapters and topics

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