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The line joining the points to intersection of the circle \mathrm{x^2+y^2=a^2} and the line \mathrm{l x+m y+n=0} subtends right angle at the origin, then
 

Option: 1

\mathrm{a^2\left(l^2+m^2\right)=n^2}

 


Option: 2

\mathrm{a^2\left(l^2+m^2\right)=2 n^2}
 


Option: 3

\mathrm{a^2\left(l^2+n^2\right)=m^2}
 


Option: 4

\mathrm{l^2+m^2+n^2=a^2}


Answers (1)

best_answer

Combined equation of \mathrm{O A \: and\: O B} are obtained by making homogeneous of circle equation with line equation \mathrm{n^2\left(x^2+y^2\right)=a^2(l x+m y)^2}

Since \mathrm{O A \perp O B}

\therefore coefficient of \mathrm{x^2+} coefficient of \mathrm{y^2=0}

\mathrm{ \Rightarrow\left(n^2-a^2 l^2\right)+\left(n^2-a^2 m^2\right)=0 }

\mathrm{\Rightarrow 2 n^2=a^2\left(l^2+m^2\right)}

Hence option 2 is correct.

Posted by

shivangi.shekhar

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