If the lines $a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$ cut the coordinates axes in concyclic points, then

a_{1} b_{1} = a_{2} b_{2}

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

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a_{1} + a_{2} = b_{1} + b_{2}

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a_{1} a_{2} = b_{1} b_{2}

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Solution

The correct option is D $a_{1} a_{2} = b_{1} b_{2}$
Let the given lines be $L_{1} \equiv a_{1} x + b_{1} y + c_{1} = 0$ and $L_{2} \equiv a_{2} x + b_{2} y + c_{2} = 0$ . Suppose $L_{1}$ meets the coordinate axes at A and B and $L_{2}$ meets at C and D. Then coordinates of A, B, C, D are
$A (- \frac{c_{1}}{a_{1}}, 0), B (0, - \frac{c_{1}}{b_{1}}) C (- \frac{c_{2}}{a_{2}}, 0), and D (0, - \frac{c_{2}}{b_{2}})$
Since A, B, C, D are concyclic, therefore OA.OC=OD.OB

$\Rightarrow (- \frac{c_{1}}{a_{1}}) (- \frac{c_{2}}{a_{2}}) = (\frac{c_{2}}{b_{2}}) (- \frac{c_{1}}{b_{1}}) \Rightarrow a_{1} a_{2} = b_{1} b_{2}$

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Similar questions

a_{1} x + b_{1} y + c_{1} = 0

a_{2} x + b_{2} y + c_{2} = 0

If the lines $a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$ cut the coordinates axes in concyclic points, then

a_{1} x + b_{1} y + c_{1} = 0

a_{2} x + b_{2} y + c_{2} = 0

| a_{1} a_{2} | = b_{1} b_{2} |

a_{1} x + b_{1} y + c_{1} = 0 a n d a_{2} x + b_{2} y + c_{2} = 0

a_{1} a_{2} = b_{1} b_{2}

(A);

(a_{1} x + b_{1} y + c_{1}) + (a_{2} x + b_{2} y + c_{2}) + a_{3} x + b_{3} y + c_{3}) = 0,

a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0, a_{3} x + b_{3} y + c_{3} = 0

0,

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