If the line $a_{1} x b_{1} y + c_{1} = 0$ and $a_{2} x b_{2} y + c_{2} = 0$ cut the coordinate axis in concyclic point, 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 $a_{1} a_{2} = b_{1} b_{2}$
The concyclic points are those which lie on same circle.

Let the equation of cirlce formed by joining those four points be
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ ....(1)

The intersection points of $a_{1} x + b_{1} y + c_{1} = 0$ with coordinate axes are
$(\frac{- c_{1}}{a_{1}}, 0), (0, \frac{- c_{1}}{b_{1}})$

And, intersection points of $a_{2} x + b_{2} y + c_{2} = 0$ with coordinate axes are
$(\frac{- c_{2}}{a_{2}}, 0), (0, \frac{- c_{2}}{b_{2}})$

Putting y = 0 in (1),
we get $x^{2} + 2 g x + c = 0$ whose roots must be $(\frac{- c_{1}}{a_{1}}, \frac{- c_{2}}{a_{2}})$
$\to c = \frac{c_{1} c_{2}}{a_{1} a_{2}}$ ....(2)

Similarly, putting x = 0 in (1),
$\to c = \frac{c_{1} c_{2}}{b_{1} b_{2}}$ ....(3)

Using 2 and 3,
$a_{1} a_{2} = b_{1} b_{2}$

Hence, (D) is the correct answer.

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