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If two lines a a_1 x+b_1 y+c_1=0 and \mathrm{a_2 x+b_2 y+c_2=0}cut the co-ordinate axes in concyclic points then

 

Option: 1

\mathrm{a_{1}a_{2}=b_{1}b_{2}}


Option: 2

\mathrm{\frac{a_1}{b_1}=\frac{a_2}{b_2}}


Option: 3

\mathrm{a_1 b_1=a_2 b_2}


Option: 4

Nome of these


Answers (1)

best_answer

  

Co-ordinate of A and B are \mathrm{\left(-\frac{c_1}{a_1}, 0\right) \text { and }\left(0,-\frac{c_1}{b_1}\right)} 

co-ordinates of C and D are \mathrm{\left(-\frac{\mathrm{c}_2}{\mathrm{a}_2}, 0\right) \text { and }\left(0,-\frac{\mathrm{c}_2}{\mathrm{~b}_2}\right)}

Equation of circle through AB is

\mathrm{\left(x+\frac{c_1}{a_1}\right)(x-0)+(y-0)\left(y+\frac{c_1}{b_1}\right)+\lambda\left(a_1 x+b_1 y+c_1\right)}

Or,

\mathrm{\lambda=\frac{x\left(x+\frac{c_1}{a_1}\right)+y\left(y+\frac{c_1}{b_1}\right)}{-\left(a_1 x+b_1 y+c_1\right)}}

It passes through  \mathrm{\left(-\frac{\mathrm{c}_2}{\mathrm{a}_2}, 0\right)}

\mathrm{\lambda=\frac{-\frac{c_2}{a_2}\left(\frac{c_1}{a_1}-\frac{c_2}{a_2}\right)}{-\left(a_1-\frac{c_2}{a_2}+c_1\right)}=\frac{c_2\left(c_1 a_2-c_2 a_1\right)}{a_1 a_2\left(c_1 a_2-c_2 a_1\right)}=\frac{c_2}{a_1 a_2}}

It passes through \mathrm{\left(0,-\frac{c_2}{b_2}\right)}

\mathrm{\lambda=\frac{-\frac{c_2}{b_2}\left(\frac{c_1}{b_1}-\frac{c_2}{b_2}\right)}{-\left(b_1 \frac{-c_2}{b_2}+c_1\right)}=\frac{c_2\left(c_1 b_2-c_2 b_1\right)}{b_1 b_2\left(c_1 b_2-c_2 b_1\right)}=\frac{c_2}{b_1 b_2}}

Hence from  \mathrm{\frac{c_2}{a_1 a_2}=\frac{c_2}{b_1 b_2}}

or                 \mathrm{a_1 a_2=b_1 b_2}

 

 

Posted by

Gaurav

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