Question

If two lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ cut the coordinate axes in concyclic points,then

• A. $a_1{a}_2+b_1{b}_2=0$
• B. $a_1{a}_2-b_1{b}_2=0$
• C. $a_1{b}_1+a_2{b}_2=0$
• D. $a_1{b}_1-a_2{b}_2=0$

Answer 1

The correct option is B $a_1{a}_2-b_1{b}_2=0$

Given that, two lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ cut the coordinate axes in concyclic points.
Let, $S=0$ be the equation of circle passing through those four points.
$a_{1}x+b_{1}y+c_1=0$ cuts coordinate axes at $A\left(\frac{-c_1}{a_1},0\right)$ and $B(0,\frac{-c_1}{b_1})$ and
$a_{2}x+b_{2}y+c_2=0$ cuts coordinate axes at $C\left(\frac{-c_2}{a_2},0\right)$ and $D(0,\frac{-c_2}{b_2})$
If $P(0,0)$ is the exterior point from which line are drawn to cut $S=0$ at $A,C$ and $B,D$
$\Rightarrow PA.PC=PB.PD=S_{11}$
$\Rightarrow \frac{c_1{c}_2}{a_{1}1a_2}=\frac{c_1{c}_2}{b_1{b}_2}$
$\Rightarrow a_1{a}_2=b_1{b}_2$
$\therefore a_1{a}_2-b_1{b}_2=0$
Hence, option B.