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  • If two lines <img alt="\mathrm{a_1 x+b_1 y+c_1=0 \: and \: a_2 x+b_2 y+c_2=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Ba_1%20x&plus;b_1%20y&plus;c_1%3D0%20%5C%3A%20

If two lines \mathrm{a_1 x+b_1 y+c_1=0 \: and \: 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{\mathrm{a}_1}{\mathrm{~b}_1}=\frac{\mathrm{a}_2}{\mathrm{~b}_2}}
 


Option: 3

\mathrm{a_1 b_1=a_2 b_2}
 


Option: 4

None of these


Answers (1)

best_answer

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

co-ordinates of  \mathrm{C \text { and } D \text { are }\left(-\frac{c_2}{a_2}, 0\right) \text { and }\left(0,-\frac{c_2}{b_2}\right)}

Equation of circle through \mathrm{A B} 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)=0}

\mathrm{ \text { or } \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{c_2}{a_2}, 0\right) } so

\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{\mathrm{c}_2}{\mathrm{~b}_2}\right) } so

\mathrm{ \lambda=\quad \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} }

\mathrm{ or \: \: a_1 a_2=b_1 b_2 }

Alternate

Since A, B, C and D we concyclic points.

So,

\mathrm{ O A . O C=O B . O D }

\mathrm{ \frac{-c_1}{a_1} \times \frac{c_2}{a_2}=\frac{-c_1}{b_1} \times \frac{c_2}{b_2} }

\mathrm{ \Rightarrow a_1 a_2=b_1 b_2 . }

Hence option 1 is correct.






 

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