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The chords of  contact of  the  pair  of  tangents  drawn from each point  on the  line 
\mathrm{2 x+y=4} to the circle\mathrm{x^2+y^2=1}  pass through  fixed  point

 

Option: 1

(2,4)


Option: 2

\left(-\frac{1}{2},-\frac{1}{4}\right)


Option: 3

\left(\frac{1}{2},\frac{1}{4}\right)


Option: 4

(-2,-4)


Answers (1)

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The  chord  of  contact of  tangents  from \mathrm{\(\alpha, \beta) } is 

\mathrm{\alpha x+\beta y=1}        .... [1]

\mathrm{\text { Also, }(\alpha, \beta) \text { lies on } 2 x+y=4 \text {, so that } 2 \alpha+\beta=4}

\mathrm{\Rightarrow \frac{\alpha}{2}+\frac{\beta}{4}=1 }

Hence, (1) passes through \mathrm{\left(\frac{1}{2}, \frac{1}{4}\right) }

 

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jitender.kumar

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