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Tangents are drawn to the circle \mathrm{x^{2}+y^{2}=1} at the points where it is met by the circles, \mathrm{x^{2}+y^{2}-(\lambda+6) x+(8-2 \lambda) y-3=0 . \lambda} being the variable. The locus of the point of intersection of these tangents is :

Option: 1

\mathrm{2 x-y+10=0}


Option: 2

\mathrm{x}+2 \mathrm{y}-10=0


Option: 3

\mathrm{x-2 y+10=0}


Option: 4

2 \mathrm{x}+\mathrm{y}-10=0


Answers (1)

best_answer

[Hint: compare chord of contact of the pair of tangents from \left(x_{1}, y_{1}\right) to the circle x^{2}+y^{2}=1 with the common chord between the two circles and eliminate \lambda ]

Locus of point of intersection of tangents

chord of contact of \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ w.r.t. $\quad \mathrm{x}^{2}+\mathrm{y}^{2}=1$ is $\mathrm{xx}_{1}+\mathrm{yy}_{1}=1\:\mathrm{ \left ( AB \right )} \cdots(1)

\mathrm{AB}  is also common chord between two circles

\mathrm{\therefore \quad-1+(\lambda+6) \mathrm{x}-(8-2 \lambda) \mathrm{y}+3=0}
\mathrm{\Rightarrow \quad(\lambda+6) \mathrm{x}-(8-2 \lambda) \mathrm{y}+2=0}\quad \cdots(2)

comparing (1) and (2) we get
\mathrm{\frac{x_{1}}{\lambda+6}=\frac{y}{2 \lambda-8}=\frac{-1}{2}}

eliminate \mathrm{\lambda \Rightarrow \quad 2 \mathrm{x}-\mathrm{y}+10=0}

Posted by

manish painkra

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