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  • Tangents are drawn to the circle <img alt="\mathrm{x}^{2}+\mathrm{y}^{2}=1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%7D%5E%7B2%7D&plus;%5Cmathrm%7By%7D%5E%7B2%7D%3D1"

Tangents are drawn to the circle \mathrm{x}^{2}+\mathrm{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 }. \mathrm{\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 y-10=0}


Option: 3

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


Option: 4

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


Answers (1)

best_answer

 Locus of point of intersection of tangents

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

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}}

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


 

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Rishi

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